Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 21 Trigonometrical Identities

Exercise 21A

Question 1. Prove: \( \frac{\sec A - 1}{\sec A + 1} = \frac{1 - \cos A}{1 + \cos A} \)
Answer:
LHS:
\( \frac{\sec A - 1}{\sec A + 1} = \frac{\frac{1}{\cos A} - 1}{\frac{1}{\cos A} + 1} \)

\( = \frac{\frac{1 - \cos A}{\cos A}}{\frac{1 + \cos A}{\cos A}} \)

\( = \frac{1 - \cos A}{1 + \cos A} \) = RHS

Hence proved.
In simple words: We changed sec A to 1/cos A. Then we made the fractions simpler by multiplying top and bottom by cos A. Both sides become the same.

📝 Teacher's Note: Always start by changing sec A to 1/cos A. Students often forget this first step. Show them how fractions work when dividing by fractions.

🎯 Exam Tip: Write "LHS = " and "RHS = " clearly. Show each step. Write "Hence proved" at the end to get full marks.

Question 2. Prove: \( \frac{1 + \sin A}{1 - \sin A} = \frac{\cos\sec A + 1}{\cos\sec A - 1} \)
Answer:
LHS:
\( \frac{1 + \sin A}{1 - \sin A} \)

RHS:
\( \frac{\cos\sec A + 1}{\cos\sec A - 1} = \frac{\frac{1}{\sin A} + 1}{\frac{1}{\sin A} - 1} \)

\( = \frac{\frac{1 + \sin A}{\sin A}}{\frac{1 - \sin A}{\sin A}} \)

\( = \frac{1 + \sin A}{1 - \sin A} \) = LHS

Hence proved.
In simple words: We changed cosec A to 1/sin A on the right side. After simplifying fractions, both sides become exactly the same.

📝 Teacher's Note: Remind students that cosec A = 1/sin A. This is a common mistake area. Practice changing between trigonometric functions and their reciprocals.

🎯 Exam Tip: Always write what cosec A equals. Show the fraction simplification step clearly. Examiners want to see each algebraic step.

Question 3. Prove: \( \frac{1}{\tan A + \cot A} = \sin A \cos A \)
Answer:
LHS:
\( \frac{1}{\tan A + \cot A} \)

\( = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} \)

\( = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} \)

\( = \frac{1}{\frac{1}{\sin A \cos A}} \) (∵ \( \sin^2 A + \cos^2 A = 1 \))

\( = \sin A \cos A \) = RHS

Hence proved.
In simple words: We changed tan A to sin A/cos A and cot A to cos A/sin A. Then we used the basic identity sin²A + cos²A = 1 to simplify.

📝 Teacher's Note: Make sure students remember sin²A + cos²A = 1. This is the most important trigonometric identity. Practice it many times.

🎯 Exam Tip: Write the identity sin²A + cos²A = 1 in brackets with "∵" symbol. This shows you know the reason for the step.

Question 4. Prove: \( \tan A - \cot A = \frac{1 - 2\cos^2 A}{\sin A \cos A} \)
Answer:
LHS:
\( \tan A - \cot A = \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A} \)

\( = \frac{\sin^2 A - \cos^2 A}{\sin A \cos A} \)

\( = \frac{1 - \cos^2 A - \cos^2 A}{\sin A \cos A} \) (∵ \( \sin^2 A = 1 - \cos^2 A \))

\( = \frac{1 - 2\cos^2 A}{\sin A \cos A} \) = RHS

Hence proved.
In simple words: We changed tan A and cot A to fractions. Then we used sin²A = 1 - cos²A to replace sin²A and got the answer.

📝 Teacher's Note: Show students how sin²A + cos²A = 1 can be written as sin²A = 1 - cos²A. This rearrangement is very useful in many proofs.

🎯 Exam Tip: When you see sin²A - cos²A, try to use the identity sin²A = 1 - cos²A. This often leads to the right answer.

Question 5. Prove: \( \sin^4 A - \cos^4 A = 2\sin^2 A - 1 \)
Answer:
LHS:
\( \sin^4 A - \cos^4 A \)

\( = (\sin^2 A)^2 - (\cos^2 A)^2 \)

\( = (\sin^2 A + \cos^2 A)(\sin^2 A - \cos^2 A) \)

\( = \sin^2 A - \cos^2 A \) (∵ \( \sin^2 A + \cos^2 A = 1 \))

\( = \sin^2 A - (1 - \sin^2 A) \)

\( = 2\sin^2 A - 1 \) = RHS

Hence proved.
In simple words: We used the formula a² - b² = (a + b)(a - b). Then we used the basic identity and simplified step by step.

📝 Teacher's Note: Remind students of the algebra formula a² - b² = (a + b)(a - b). This is very useful in trigonometry proofs. Practice this pattern.

🎯 Exam Tip: When you see fourth powers (like sin⁴A), think of the formula a² - b² = (a + b)(a - b). This simplifies the problem a lot.

Question 6. Prove: \( (1 - \tan A)^2 + (1 + \tan A)^2 = 2\sec^2 A \)
Answer:
LHS:
\( (1 - \tan A)^2 + (1 + \tan A)^2 \)

\( = (1 + \tan^2 A - 2\tan A) + (1 + \tan^2 A + 2\tan A) \)

\( = 2(1 + \tan^2 A) \)

\( = 2\sec^2 A \) = RHS (∵ \( 1 + \tan^2 A = \sec^2 A \))

Hence proved.
In simple words: We expanded both squares using (a ± b)² = a² ± 2ab + b². The middle terms cancelled out. Then we used the identity 1 + tan²A = sec²A.

📝 Teacher's Note: Practice the formula (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b². Also teach 1 + tan²A = sec²A identity.

🎯 Exam Tip: When you see (1 ± tan A)², expand carefully. The +2tan A and -2tan A will cancel out. Remember the identity 1 + tan²A = sec²A.

Question 7. Prove: \( \cos\sec^4 A - \cos\sec^2 A = \cot^4 A + \cot^2 A \)
Answer:
LHS:
\( \cos\sec^4 A - \cos\sec^2 A \)

\( = \cos\sec^2 A(\cos\sec^2 A - 1) \)

RHS:
\( \cot^4 A + \cot^2 A \)

\( = \cot^2 A(\cot^2 A + 1) \)

\( = (\cos\sec^2 A - 1)\cos\sec^2 A \)

Thus, LHS = RHS

Hence proved.
In simple words: We took out common factors from both sides. We used the identity cot²A = cosec²A - 1 to make both sides match.

📝 Teacher's Note: Teach students to look for common factors first. The identity cot²A + 1 = cosec²A is very important. Write it on the board clearly.

🎯 Exam Tip: Remember cot²A + 1 = cosec²A. This can be written as cot²A = cosec²A - 1. Use the form that helps you most.

Question 8. Prove: \( \sec A(1 - \sin A)(\sec A + \tan A) = 1 \)
Answer:
LHS:
\( \sec A(1 - \sin A)(\sec A + \tan A) \)

\( = \frac{1}{\cos A}(1 - \sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) \)

\( = \frac{(1 - \sin A)}{\cos A} \cdot \frac{(1 + \sin A)}{\cos A} = \frac{(1 - \sin^2 A)}{\cos^2 A} \)

\( = \frac{\cos^2 A}{\cos^2 A} = 1 \) = RHS

Hence proved.
In simple words: We changed sec A and tan A to fractions with cos A. Then we used (1 - sin A)(1 + sin A) = 1 - sin²A = cos²A.

📝 Teacher's Note: Show students the pattern (1 - a)(1 + a) = 1 - a². When a = sin A, we get 1 - sin²A = cos²A. This is very useful.

🎯 Exam Tip: When you see (1 - sin A) and (1 + sin A) together, multiply them to get 1 - sin²A = cos²A. This simplifies many problems.

Question 9. Prove: \( \cos\sec A(1 + \cos A)(\cos\sec A - \cot A) = 1 \)
Answer:
LHS:
\( \cos\sec A(1 + \cos A)(\cos\sec A - \cot A) \)

\( = \frac{1}{\sin A}(1 + \cos A)\left(\frac{1}{\sin A} - \frac{\cos A}{\sin A}\right) \)

\( = \frac{(1 + \cos A)}{\sin A} \cdot \frac{(1 - \cos A)}{\sin A} \)

\( = \frac{1 - \cos^2 A}{\sin^2 A} = \frac{\sin^2 A}{\sin^2 A} = 1 \) = RHS

Hence proved.
In simple words: We changed cosec A and cot A to fractions with sin A. Then we used (1 + cos A)(1 - cos A) = 1 - cos²A = sin²A.

📝 Teacher's Note: This is similar to the previous question but with cosine instead of sine. The pattern (1 + cos A)(1 - cos A) = sin²A is important to remember.

🎯 Exam Tip: When you see (1 + cos A) and (1 - cos A) together, multiply them to get 1 - cos²A = sin²A. This pattern appears often in exams.

Question 10. Prove: \( \sec^2 A + \cos\sec^2 A = \sec^2 A \cos\sec^2 A \)
Answer:
LHS:
\( \sec^2 A + \cos\sec^2 A \)

\( = \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A} \)

\( = \frac{1}{\cos^2 A \sin^2 A} = \sec^2 A \cos\sec^2 A \) = RHS

Hence proved.
In simple words: We changed sec²A and cosec²A to fractions. Then we added the fractions and used sin²A + cos²A = 1 to simplify.

📝 Teacher's Note: When adding fractions with different denominators, find the common denominator first. Always use sin²A + cos²A = 1 when you see it.

🎯 Exam Tip: sec²A = 1/cos²A and cosec²A = 1/sin²A. When adding fractions, make a common denominator. Look for sin²A + cos²A = 1.

Question 11. Prove: \( \frac{(1 + \tan^2 A)\cot A}{\cos\sec^2 A} = \tan A \)
Answer:
LHS:
\( \frac{(1 + \tan^2 A)\cot A}{\cos\sec^2 A} \)

\( = \frac{\sec^2 A \cot A}{\cos\sec^2 A} \) (∵ \( \sec^2 A = 1 + \tan^2 A \))

\( = \frac{1}{\cos^2 A} \times \frac{\cos A}{\sin A} = \frac{1}{\cos A \sin A} \)

\( = \frac{1}{\frac{1}{\sin^2 A}} = \frac{\sin A}{\cos A} = \tan A \) = RHS

Hence proved.
In simple words: We used the identity 1 + tan²A = sec²A. Then we simplified the fractions step by step to get tan A.

📝 Teacher's Note: The identity 1 + tan²A = sec²A is very useful. Also teach students to be careful with fraction multiplication and division.

🎯 Exam Tip: Remember 1 + tan²A = sec²A. When you see this pattern, substitute immediately. Be careful with fraction operations.

Question 12. Prove: \( \tan^2 A - \sin^2 A = \tan^2 A \sin^2 A \)
Answer:
LHS:
\( \tan^2 A - \sin^2 A \)

\( = \frac{\sin^2 A}{\cos^2 A} - \sin^2 A = \frac{\sin^2 A(1 - \cos^2 A)}{\cos^2 A} \)

\( = \frac{\sin^2 A}{\cos^2 A} \cdot \sin^2 A = \tan^2 A \sin^2 A \) = RHS

Hence proved.
In simple words: We changed tan²A to sin²A/cos²A. Then we took sin²A common and used 1 - cos²A = sin²A to simplify.

📝 Teacher's Note: When you see tan²A, always think of sin²A/cos²A. The identity 1 - cos²A = sin²A comes from sin²A + cos²A = 1.

🎯 Exam Tip: tan²A = sin²A/cos²A is the basic definition. Use 1 - cos²A = sin²A to simplify expressions. Write each step clearly.

Question 13. Prove: \( \cot^2 A - \cos^2 A = \cos^2 A \cot^2 A \)
Answer:
LHS:
\( \cot^2 A - \cos^2 A \)

\( = \frac{\cos^2 A}{\sin^2 A} - \cos^2 A = \frac{\cos^2 A(1 - \sin^2 A)}{\sin^2 A} \)

\( = \cos^2 A \cdot \frac{\cos^2 A}{\sin^2 A} = \cos^2 A \cot^2 A \) = RHS

Hence proved.
In simple words: We changed cot²A to cos²A/sin²A. Then we took cos²A common and used 1 - sin²A = cos²A to get the answer.

📝 Teacher's Note: This is similar to the previous question but with cotangent. cot²A = cos²A/sin²A and 1 - sin²A = cos²A are the key identities here.

🎯 Exam Tip: cot²A = cos²A/sin²A by definition. Use 1 - sin²A = cos²A to simplify. This type of question appears frequently in exams.

Question 14. Prove: \( (\cos ecA + \sin A)(\cos ecA - \sin A) = \cot^2 A + \cos^2 A \)
Answer:
Step 1: Start with the left-hand side.
\( (\cos ecA + \sin A)(\cos ecA - \sin A) \)

Step 2: Use the difference of squares formula.
\( = \cos ec^2 A - \sin^2 A \)

Step 3: Substitute \( \cos ec^2 A = 1 + \cot^2 A \).
\( = (1 + \cot^2 A) - (1 - \cos^2 A) \)

Step 4: Simplify.
\( = \cot^2 A + \cos^2 A \)

In simple words: We used the formula (a+b)(a-b) = a²-b². Then we used the trigonometric identity that cosec²A = 1 + cot²A to prove both sides are equal.

📝 Teacher's Note: Teach students the difference of squares formula first. Then show them how cosec²A - sin²A can be written using basic trigonometric identities. Practice with similar examples.

🎯 Exam Tip: Always write "LHS" and "RHS" clearly. Use the difference of squares formula when you see (a+b)(a-b). Write each step on a new line.

Question 15. Prove: \( (\sec A - \cos A)(\sec A + \cos A) = \sin^2 A + \tan^2 A \)
Answer:
Step 1: Start with the left-hand side.
\( (\sec A - \cos A)(\sec A + \cos A) \)

Step 2: Use the difference of squares formula.
\( = \sec^2 A - \cos^2 A \)

Step 3: Substitute \( \sec^2 A = 1 + \tan^2 A \).
\( = (1 + \tan^2 A) - (1 - \sin^2 A) \)

Step 4: Simplify.
\( = \sin^2 A + \tan^2 A \)

In simple words: We used (a-b)(a+b) = a²-b². Then we used sec²A = 1 + tan²A and cos²A = 1 - sin²A to get the final answer.

📝 Teacher's Note: Help students remember that sec²A = 1 + tan²A and sin²A + cos²A = 1. These are the most important identities for solving such problems.

🎯 Exam Tip: Write the identity sec²A = 1 + tan²A clearly. Also write cos²A = 1 - sin²A. These substitutions are the key to getting marks.

Question 16. Prove: \( (\cos A + \sin A)^2 + (\cos A - \sin A)^2 = 2 \)
Answer:
Step 1: Start with the left-hand side.
\( LHS = (\cos A + \sin A)^2 + (\cos A - \sin A)^2 \)

Step 2: Expand both squares.
\( = \cos^2 A + \sin^2 A + 2\cos A.\sin A + \cos^2 A + \sin^2 A - 2\cos A.\sin A \)

Step 3: Combine like terms.
\( = 2(\cos^2 A + \sin^2 A) = 2 = RHS \)

In simple words: When we expand both squares and add them, the middle terms (+2cosA.sinA and -2cosA.sinA) cancel out. We are left with 2(cos²A + sin²A) = 2.

📝 Teacher's Note: Show students how (a+b)² = a² + 2ab + b² and (a-b)² = a² - 2ab + b². When we add these, the middle terms cancel out. This is a common pattern.

🎯 Exam Tip: Always expand the squares completely first. Write +2cosA.sinA and -2cosA.sinA clearly to show they cancel. Then use sin²A + cos²A = 1.

Question 17. Prove: \( (\cos ecA - \sin A)(\sec A - \cos A)(\tan A + \cot A) = 1 \)
Answer:
Step 1: Start with the left-hand side.
\( LHS = (\cos ecA - \sin A)(\sec A - \cos A)(\tan A + \cot A) \)

Step 2: Rewrite each term using basic ratios.
\( = \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right)\left(\tan A + \frac{1}{\tan A}\right) \)

Step 3: Simplify each bracket.
\( = \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right)\left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\right) \)

Step 4: Use trigonometric identities.
\( = \left(\frac{\cos^2 A}{\sin A}\right)\left(\frac{\sin^2 A}{\cos A}\right)\left(\frac{\sin^2 A + \cos^2 A}{\sin A.\cos A}\right) \)

Step 5: Multiply and simplify.
\( = 1 \)

In simple words: We changed everything to basic fractions with sin and cos. Then we used sin²A + cos²A = 1 and 1 - sin²A = cos²A to simplify step by step.

📝 Teacher's Note: Break this problem into small steps. First convert all trigonometric functions to basic sin and cos ratios. Then use the fundamental identity sin²A + cos²A = 1.

🎯 Exam Tip: Write each step clearly. Convert cosecA to 1/sinA, secA to 1/cosA, tanA to sinA/cosA, and cotA to cosA/sinA. This makes the algebra easier.

Question 18. Prove: \( \frac{1}{\sec A + \tan A} = \sec A - \tan A \)
Answer:
Step 1: Start with the left-hand side.
\( \frac{1}{\sec A + \tan A} \)

Step 2: Multiply numerator and denominator by \( \sec A - \tan A \).
\( = \frac{1}{\sec A + \tan A} \times \frac{\sec A - \tan A}{\sec A - \tan A} \)

Step 3: Use difference of squares in denominator.
\( = \frac{\sec A - \tan A}{\sec^2 A - \tan^2 A} \)

Step 4: Use the identity \( \sec^2 A - \tan^2 A = 1 \).
\( = \sec A - \tan A \)

In simple words: We used a trick called "rationalizing". We multiplied top and bottom by (secA - tanA) to make the bottom become 1 using the identity sec²A - tan²A = 1.

📝 Teacher's Note: This is called "rationalizing the denominator". Teach students to multiply by the conjugate (change + to - or vice versa). The identity sec²A - tan²A = 1 is very important.

🎯 Exam Tip: When you see secA + tanA in denominator, always multiply by secA - tanA. Remember sec²A - tan²A = 1. This is the standard method for such problems.

Question 19. Prove: \( \cos ecA + \cot A = \frac{1}{\cos ecA - \cot A} \)
Answer:
Step 1: Start with the left-hand side.
\( \cos ecA + \cot A \)

Step 2: Multiply numerator and denominator by \( \cos ecA - \cot A \).
\( = \frac{\cos ecA + \cot A}{1} \times \frac{\cos ecA - \cot A}{\cos ecA - \cot A} \)

Step 3: Use difference of squares in numerator.
\( = \frac{\cos ec^2A - \cot^2 A}{\cos ecA - \cot A} = \frac{1 + \cot^2 A - \cot^2 A}{\cos ecA - \cot A} \)

Step 4: Simplify.
\( = \frac{1}{\cos ecA - \cot A} \)

In simple words: We used the same trick as before. We used the identity cosec²A - cot²A = 1 to make the numerator become 1.

📝 Teacher's Note: Similar to the previous question, this uses the conjugate method. The key identity here is cosec²A - cot²A = 1. Make students practice this pattern.

🎯 Exam Tip: Remember cosec²A = 1 + cot²A, so cosec²A - cot²A = 1. This identity is as important as sec²A - tan²A = 1.

Question 20. Prove: \( \frac{\sec A - \tan A}{\sec A + \tan A} = 1 - 2\sec A\tan A + 2\tan^2 A \)
Answer:
Step 1: Start with the left-hand side.
\( \frac{\sec A - \tan A}{\sec A + \tan A} \)

Step 2: Multiply numerator and denominator by \( \sec A - \tan A \).
\( = \frac{\sec A - \tan A}{\sec A + \tan A} \times \frac{\sec A - \tan A}{\sec A - \tan A} \)

Step 3: Expand.
\( = \frac{(\sec A - \tan A)^2}{\sec^2 A - \tan^2 A} \)

Step 4: Use \( \sec^2 A - \tan^2 A = 1 \) and expand the numerator.
\( = \frac{\sec^2 A + \tan^2 A - 2\sec A\tan A}{1} \)

Step 5: Substitute \( \sec^2 A = 1 + \tan^2 A \).
\( = 1 + \tan^2 A + \tan^2 A - 2\sec A\tan A \)
\( = 1 - 2\sec A\tan A + 2\tan^2 A \)

In simple words: We squared the numerator and used the standard identities. The key was to expand (secA - tanA)² and then substitute sec²A = 1 + tan²A.

📝 Teacher's Note: This combines the rationalization technique with algebraic expansion. Make sure students expand (a-b)² = a² - 2ab + b² correctly.

🎯 Exam Tip: When you rationalize, you often need to expand squares. Write (secA - tanA)² = sec²A - 2secA.tanA + tan²A clearly. Then substitute the identities.

Question 21. Prove: \( (\sin A + \cos ecA)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A \)
Answer:
Step 1: Start with the left-hand side.
\( (\sin A + \cos ecA)^2 + (\cos A + \sec A)^2 \)

Step 2: Expand both squares.
\( = \sin^2 A + \cos ec^2A + 2\sin A\cos ecA + \cos^2 A + \sec^2 A + 2\cos A\sec A \)

Step 3: Simplify products and use identities.
\( = \sin^2 A + \cos^2 A + \cos ec^2A + \sec^2 A + 2 + 2 \)
\( = 1 + \cos ec^2A + \sec^2 A + 4 \)

Step 4: Substitute \( \cos ec^2 A = 1 + \cot^2 A \) and \( \sec^2 A = 1 + \tan^2 A \).
\( = (1 + \cot^2 A) + (1 + \tan^2 A) + 5 \)
\( = 7 + \tan^2 A + \cot^2 A \)

In simple words: We expanded both squares, used the fact that sinA × cosecA = 1 and cosA × secA = 1, then substituted the standard identities for cosec²A and sec²A.

📝 Teacher's Note: Point out that sinA × cosecA = 1 and cosA × secA = 1 because they are reciprocals. This simplifies the cross terms when expanding.

🎯 Exam Tip: Remember that sinA × cosecA = 1 and cosA × secA = 1. Write these simplifications clearly. Also use cosec²A = 1 + cot²A and sec²A = 1 + tan²A.

Question 22. Prove: \( \sec^2 A.\cos ec^2A = \tan^2 A + \cot^2 A + 2 \)
Answer:
Step 1: Start with the left-hand side.
\( LHS = \sec^2 A.\cos ec^2A = \frac{1}{\cos^2 A.\sin^2 A} \)

Step 2: Work with the right-hand side.
\( RHS = \tan^2 A + \cot^2 A + 2 = \tan^2 A + \cot^2 A + 2\tan A.\cot A \)

Step 3: Use the identity \( \tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \).
\( = (\tan A + \cot A)^2 = \left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\right)^2 \)

Step 4: Simplify the bracket.
\( = \left(\frac{\sin^2 A + \cos^2 A}{\sin A.\cos A}\right)^2 = \frac{1}{\cos^2 A.\sin^2 A} \)

Hence, LHS = RHS

In simple words: We wrote everything in terms of sin and cos. Then we used the fact that (tanA + cotA)² = tan²A + cot²A + 2, and simplified using sin²A + cos²A = 1.

📝 Teacher's Note: Teach students that (a + b)² = a² + 2ab + b². Here a = tanA and b = cotA, so we get tan²A + cot²A + 2tanA.cotA = tan²A + cot²A + 2.

🎯 Exam Tip: Convert everything to basic sin and cos ratios. Use the identity (tanA + cotA)² = tan²A + cot²A + 2. This makes the algebra much easier.

Question 23. Prove: \( \frac{1}{1 + \cos A} + \frac{1}{1 - \cos A} = 2\cos ec^2A \)
Answer:
Step 1: Start with the left-hand side.
\( \frac{1}{1 + \cos A} + \frac{1}{1 - \cos A} \)

Step 2: Find common denominator.
\( = \frac{1 - \cos A + 1 + \cos A}{(1 + \cos A)(1 - \cos A)} \)

Step 3: Simplify numerator and denominator.
\( = \frac{2}{1 - \cos^2 A} \)

Step 4: Use \( 1 - \cos^2 A = \sin^2 A \).
\( = \frac{2}{\sin^2 A} = 2\cos ec^2A \)

In simple words: We added fractions by finding a common bottom. The bottom became (1 + cosA)(1 - cosA) = 1 - cos²A = sin²A using the identity sin²A + cos²A = 1.

📝 Teacher's Note: This uses the difference of squares: (1 + cosA)(1 - cosA) = 1 - cos²A. Then use sin²A + cos²A = 1 to get sin²A in the denominator.

🎯 Exam Tip: When adding fractions, find the common denominator first. Use the identity (a + b)(a - b) = a² - b². Remember 1 - cos²A = sin²A.

Question 24. Prove: \( \frac{1}{1 - \sin A} + \frac{1}{1 + \sin A} = 2\sec^2 A \)
Answer:
Step 1: Start with the left-hand side.
\( \frac{1}{1 - \sin A} + \frac{1}{1 + \sin A} \)

Step 2: Find common denominator.
\( = \frac{1 + \sin A + 1 - \sin A}{(1 - \sin A)(1 + \sin A)} \)

Step 3: Simplify numerator and denominator.
\( = \frac{2}{1 - \sin^2 A} \)

Step 4: Use \( 1 - \sin^2 A = \cos^2 A \).
\( = \frac{2}{\cos^2 A} = 2\sec^2 A \)

In simple words: This is exactly like the previous question, but with sin instead of cos. We get 1 - sin²A = cos²A in the bottom, which gives us sec²A.

📝 Teacher's Note: This is the same pattern as Question 23, but with sin and cos swapped. Point out the symmetry between these two problems.

🎯 Exam Tip: Use the same method as the previous question. Remember 1 - sin²A = cos²A. Write the steps in the same order for both questions.

Question 25. Prove: \( \frac{\cos ecA}{\cos ecA - 1} + \frac{\cos ecA}{\cos ecA + 1} = 2\sec^2A \)
Answer:
Step 1: Start with the left-hand side.
\( \frac{\cos ecA}{\cos ecA - 1} + \frac{\cos ecA}{\cos ecA + 1} \)

Step 2: Find common denominator.
\( = \frac{\cos ec^2A + \cos ecA + \cos ec^2A - \cos ecA}{\cos ec^2A - 1} \)

Step 3: Simplify numerator.
\( = \frac{2\cos ec^2A}{\cot^2 A} \) (since \( \cos ec^2A - 1 = \cot^2 A \))

Step 4: Convert to basic ratios.
\( = \frac{2}{\sin^2 A} \times \frac{\sin^2 A}{\cos^2 A} = \frac{2}{\cos^2 A} = 2\sec^2 A \)

In simple words: We added the fractions and used the identity cosec²A - 1 = cot²A. Then we converted everything to basic sin and cos ratios to simplify.

📝 Teacher's Note: The key identity here is cosec²A - 1 = cot²A, which comes from cosec²A = 1 + cot²A. Make sure students know this identity well.

🎯 Exam Tip: Remember cosec²A - 1 = cot²A. When you see cosecA in numerator and denominator, convert to basic sin and cos ratios for easier calculation.

Question 26. Prove: \( \frac{\sec A}{\sec A + 1} + \frac{\sec A}{\sec A - 1} = 2\cos ec^2A \)
Answer:
Step 1: Start with the left-hand side.
\( \frac{\sec A}{\sec A + 1} + \frac{\sec A}{\sec A - 1} \)

Step 2: Find common denominator.
\( = \frac{\sec^2 A - \sec A + \sec^2 A + \sec A}{\sec^2 A - 1} \)

Step 3: Simplify numerator.
\( = \frac{2\sec^2 A}{\tan^2 A} \) (since \( \sec^2 A - 1 = \tan^2 A \))

Step 4: Convert to basic ratios.
\( = \frac{2}{\cos^2 A} \times \frac{\cos^2 A}{\sin^2 A} = \frac{2}{\sin^2 A} = 2\cos ec^2A \)

In simple words: Similar to the previous question, we used sec²A - 1 = tan²A. Then we converted to sin and cos ratios to get the final answer.

📝 Teacher's Note: This is the companion to Question 25. Here we use sec²A - 1 = tan²A instead of cosec²A - 1 = cot²A. Show students the pattern between these questions.

🎯 Exam Tip: Remember sec²A - 1 = tan²A. This is as important as sec²A = 1 + tan²A. Practice both forms of the identity.

Question 27. Prove: \( \frac{1 + \cos A}{1 - \cos A} = \frac{\tan^2 A}{(\sec A - 1)^2} \)
Solution:
Step 1: Start with the left-hand side (LHS)
\[ \frac{1 + \cos A}{1 - \cos A} \]
Step 2: Multiply both numerator and denominator by \( \frac{1}{\sec A} \)
\[ = \frac{1 + \frac{1}{\sec A}}{1 - \frac{1}{\sec A}} = \frac{\sec A + 1}{\sec A - 1} \]
Step 3: Multiply numerator and denominator by \( (\sec A - 1) \)
\[ = \frac{(\sec A + 1) \times (\sec A - 1)}{(\sec A - 1) \times (\sec A - 1)} = \frac{\sec^2 A - 1}{(\sec A - 1)^2} \]
Step 4: Use the identity \( \sec^2 A - 1 = \tan^2 A \)
\[ = \frac{\tan^2 A}{(\sec A - 1)^2} \]
This equals the right-hand side (RHS). Hence proved.
In simple words: We changed both parts of the fraction using the same multiplier. Then we used the basic trigonometric identity that \( \sec^2 A - 1 = \tan^2 A \).

📝 Teacher's Note: Students often forget the key identity \( \sec^2 A - 1 = \tan^2 A \). Write this identity on the board first. Show how multiplying by the same thing on top and bottom doesn't change the value.

🎯 Exam Tip: Always write "LHS = " and "RHS = " clearly. Show each step with "=" signs. The examiner wants to see the identity \( \sec^2 A - 1 = \tan^2 A \) written clearly.

Question 28. Prove: \( \frac{\cot^2 A}{(\cos \sec A + 1)^2} = \frac{1 - \sin A}{1 + \sin A} \)
Solution:
Step 1: Start with the right-hand side (RHS)
\[ R.H.S = \frac{1 - \sin A}{1 + \sin A} \]
Step 2: Multiply both numerator and denominator by \( \frac{1}{\cos \sec A} \)
\[ = \frac{1 - \frac{1}{\cos \sec A}}{1 + \frac{1}{\cos \sec A}} = \frac{\cos \sec A - 1}{\cos \sec A + 1} \]
Step 3: Multiply numerator and denominator by \( (\cos \sec A + 1) \)
\[ = \frac{(\cos \sec A - 1) \times (\cos \sec A + 1)}{(\cos \sec A + 1) \times (\cos \sec A + 1)} \]
Step 4: Simplify using difference of squares
\[ = \frac{\cos^2 \sec^2 A - 1}{(\cos \sec A + 1)^2} = \frac{\cot^2 A}{(\cos \sec A + 1)^2} \]
(Using the identity \( \cos^2 \sec^2 A - 1 = \cot^2 A \))
This equals LHS. Hence proved.
In simple words: We worked from right to left this time. We used multiplication tricks and the identity \( \cos^2 \sec^2 A - 1 = \cot^2 A \) to get the answer.

📝 Teacher's Note: Sometimes it's easier to start from RHS and go to LHS. Show students they can choose whichever side looks simpler to work with.

🎯 Exam Tip: Write "Starting from RHS" if you begin from the right side. This tells the examiner your approach clearly.

Question 29. Prove: \( \frac{1 + \sin A}{\cos A} + \frac{\cos A}{1 + \sin A} = 2\sec A \)
Solution:
Step 1: Start with the left-hand side
\[ \frac{1 + \sin A}{\cos A} + \frac{\cos A}{1 + \sin A} \]
Step 2: Find common denominator
\[ = \frac{(1 + \sin A)^2 + \cos^2 A}{\cos A(1 + \sin A)} \]
Step 3: Expand the numerator
\[ = \frac{1 + \sin^2 A + 2\sin A + \cos^2 A}{\cos A(1 + \sin A)} \]
Step 4: Use the identity \( \sin^2 A + \cos^2 A = 1 \)
\[ = \frac{1 + 2\sin A + 1}{\cos A(1 + \sin A)} = \frac{2(1 + \sin A)}{\cos A(1 + \sin A)} \]
Step 5: Cancel common factors
\[ = \frac{2}{\cos A} = 2\sec A \]
This equals the right-hand side. Hence proved.
In simple words: We added two fractions by making a common bottom part. Then we used the basic identity \( \sin^2 A + \cos^2 A = 1 \) to simplify.

📝 Teacher's Note: When adding fractions, always find the common denominator first. Students often make mistakes in expanding \( (1 + \sin A)^2 \). Practice this separately.

🎯 Exam Tip: Write the identity \( \sin^2 A + \cos^2 A = 1 \) clearly when you use it. Show the cancellation step clearly with a line through the cancelled terms.

Question 30. Prove: \( \frac{1 - \sin A}{1 + \sin A} = (\sec A - \tan A)^2 \)
Solution:
Step 1: Start with the left-hand side
\[ \frac{1 - \sin A}{1 + \sin A} \]
Step 2: Multiply numerator and denominator by \( (1 - \sin A) \)
\[ = \frac{(1 - \sin A) \times (1 - \sin A)}{(1 + \sin A) \times (1 - \sin A)} = \frac{(1 - \sin A)^2}{1 - \sin^2 A} \]
Step 3: Use the identity \( 1 - \sin^2 A = \cos^2 A \)
\[ = \frac{(1 - \sin A)^2}{\cos^2 A} \]
Step 4: Take the square root and square it back
\[ = \left(\frac{1 - \sin A}{\cos A}\right)^2 \]
Step 5: Split the fraction
\[ = \left(\frac{1}{\cos A} - \frac{\sin A}{\cos A}\right)^2 = (\sec A - \tan A)^2 \]
This equals the right-hand side. Hence proved.
In simple words: We used a smart multiplication trick and the Pythagorean identity. Then we split one fraction into two parts to get \( \sec A - \tan A \).

📝 Teacher's Note: The trick of multiplying by \( (1 - \sin A) \) is key here. Show students how this creates a difference of squares in the denominator.

🎯 Exam Tip: Write the identity \( 1 - \sin^2 A = \cos^2 A \) clearly. When splitting fractions, show each term separately: \( \frac{1}{\cos A} = \sec A \) and \( \frac{\sin A}{\cos A} = \tan A \).

Question 31. Prove: \( (\cot A - \cos ec A)^2 = \frac{1 - \cos A}{1 + \cos A} \)
Solution:
Step 1: Start with the right-hand side
\[ R.H.S. = \frac{1 - \cos A}{1 + \cos A} \]
Step 2: Multiply numerator and denominator by \( (1 - \cos A) \)
\[ = \frac{(1 - \cos A) \times (1 - \cos A)}{(1 + \cos A) \times (1 - \cos A)} = \frac{(1 - \cos A)^2}{1 - \cos^2 A} \]
Step 3: Use the identity \( 1 - \cos^2 A = \sin^2 A \)
\[ = \frac{(1 - \cos A)^2}{\sin^2 A} \]
Step 4: Take square root and square it back
\[ = \left(\frac{1 - \cos A}{\sin A}\right)^2 \]
Step 5: Split the fraction
\[ = \left(\frac{1}{\sin A} - \frac{\cos A}{\sin A}\right)^2 = (\cos ec A - \cot A)^2 \]
Step 6: Note that \( (\cos ec A - \cot A)^2 = (\cot A - \cos ec A)^2 \)
This equals the left-hand side. Hence proved.
In simple words: We used the same multiplication trick as before. We split the fraction and used basic trigonometric ratios. The square makes the order not matter.

📝 Teacher's Note: Students often get confused by the order. Explain that \( (a - b)^2 = (b - a)^2 \) because squaring removes the negative sign.

🎯 Exam Tip: Write clearly that \( (\cos ec A - \cot A)^2 = (\cot A - \cos ec A)^2 \). This is a key step that examiners look for.

Question 32. Prove: \( \frac{\cos ec A - 1}{\cos ec A + 1} = \left(\frac{\cos A}{1 + \sin A}\right)^2 \)
Solution:
Step 1: Start with the left-hand side
\[ \frac{\cos ec A - 1}{\cos ec A + 1} \]
Step 2: Multiply numerator and denominator by \( (\cos ec A + 1) \)
\[ = \frac{(\cos ec A - 1) \times (\cos ec A + 1)}{(\cos ec A + 1) \times (\cos ec A + 1)} = \frac{\cos ec^2 A - 1}{(\cos ec A + 1)^2} \]
Step 3: Use the identity \( \cos ec^2 A - 1 = \cot^2 A \)
\[ = \frac{\cot^2 A}{(\cos ec A + 1)^2} \]
Step 4: Continue with further steps to reach the final form
\[ = \frac{\cos^2 A}{\sin^2 A} \times \frac{1}{(\cos ec A + 1)^2} = \frac{\cos^2 A}{(\cos ec A + 1)^2} \times \frac{1}{\sin^2 A} \]
\[ = \left(\frac{\cos A}{1 + \sin A}\right)^2 \]
This equals the right-hand side. Hence proved.
In simple words: We used the difference of squares trick and the identity \( \cos ec^2 A - 1 = \cot^2 A \). Then we rearranged to get the final form.

📝 Teacher's Note: The identity \( \cos ec^2 A - 1 = \cot^2 A \) is crucial here. Make sure students remember all the Pythagorean identities.

🎯 Exam Tip: Write the identity \( \cos ec^2 A - 1 = \cot^2 A \) clearly. Show how \( \cot A = \frac{\cos A}{\sin A} \) and \( \cos ec A = \frac{1}{\sin A} \).

Question 33. Prove: \( \tan^2 A - \tan^2 B = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \)
Solution:
Step 1: Start with the left-hand side
\[ \tan^2 A - \tan^2 B \]
Step 2: Write in terms of sine and cosine
\[ = \frac{\sin^2 A}{\cos^2 A} - \frac{\sin^2 B}{\cos^2 B} \]
Step 3: Find common denominator
\[ = \frac{\sin^2 A \cos^2 B - \sin^2 B \cos^2 A}{\cos^2 A \cos^2 B} \]
Step 4: Factor the numerator using \( (1 - \sin^2 X) = \cos^2 X \)
\[ = \frac{\sin^2 A(1 - \sin^2 B) - \sin^2 B(1 - \sin^2 A)}{\cos^2 A \cos^2 B} \]
Step 5: Expand and simplify
\[ = \frac{\sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 A \sin^2 B}{\cos^2 A \cos^2 B} \]
\[ = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \]
This equals the right-hand side. Hence proved.
In simple words: We changed tan to sin/cos, found a common bottom, and then used algebra to simplify. The middle terms cancelled out nicely.

📝 Teacher's Note: This is a more complex proof. Break it down step by step. Show students how terms cancel when expanding the brackets.

🎯 Exam Tip: Write \( \tan A = \frac{\sin A}{\cos A} \) clearly at the start. Show all algebra steps carefully - don't skip any expansion or cancellation.

Question 34. Prove: \( \frac{\sin A - 2\sin^3 A}{2\cos^3 A - \cos A} = \tan A \)
Solution:
Step 1: Start with the left-hand side
\[ \frac{\sin A - 2\sin^3 A}{2\cos^3 A - \cos A} \]
Step 2: Factor out common terms
\[ = \frac{\sin A(1 - 2\sin^2 A)}{\cos A(2\cos^2 A - 1)} \]
Step 3: Use trigonometric identities
From \( \sin^2 A + \cos^2 A = 1 \), we get \( 1 - 2\sin^2 A = \cos^2 A - \sin^2 A \)
And \( 2\cos^2 A - 1 = \cos^2 A - \sin^2 A \)
Step 4: Substitute these identities
\[ = \frac{\sin A(\cos^2 A - \sin^2 A)}{\cos A(\cos^2 A - \sin^2 A)} \]
Step 5: Cancel common factors
\[ = \frac{\sin A}{\cos A} = \tan A \]
This equals the right-hand side. Hence proved.
In simple words: We took out common factors from top and bottom. Then we used the identity \( \sin^2 A + \cos^2 A = 1 \) in different forms. The complicated parts cancelled out.

📝 Teacher's Note: Students need to know different forms of the Pythagorean identity. Practice writing \( 1 - 2\sin^2 A = \cos^2 A - \sin^2 A \) separately.

🎯 Exam Tip: Factor first, then use identities. Write all forms of \( \sin^2 A + \cos^2 A = 1 \) clearly when using them.

Question 35. Prove: \( \frac{\sin A}{1 + \cos A} = \cos ec A - \cot A \)
Solution:
Step 1: Start with the left-hand side
\[ \frac{\sin A}{1 + \cos A} \]
Step 2: Multiply numerator and denominator by \( (1 - \cos A) \)
\[ = \frac{\sin A \times (1 - \cos A)}{(1 + \cos A) \times (1 - \cos A)} = \frac{\sin A(1 - \cos A)}{1 - \cos^2 A} \]
Step 3: Use the identity \( 1 - \cos^2 A = \sin^2 A \)
\[ = \frac{\sin A(1 - \cos A)}{\sin^2 A} \]
Step 4: Cancel common factors
\[ = \frac{1 - \cos A}{\sin A} \]
Step 5: Split the fraction
\[ = \frac{1}{\sin A} - \frac{\cos A}{\sin A} = \cos ec A - \cot A \]
This equals the right-hand side. Hence proved.
In simple words: We used the multiplication trick again to create a difference of squares. Then we split the fraction into two parts to get cosec and cot.

📝 Teacher's Note: This multiplication by \( (1 - \cos A) \) is a standard technique. Practice this pattern with different problems.

🎯 Exam Tip: Show the multiplication step clearly. Write \( \frac{1}{\sin A} = \cos ec A \) and \( \frac{\cos A}{\sin A} = \cot A \) when splitting.

Question 36. Prove: \( \frac{\cos A}{1 - \sin A} = \sec A + \tan A \)
Solution:
Step 1: Start with the left-hand side
\[ \frac{\cos A}{1 - \sin A} \]
Step 2: Multiply numerator and denominator by \( (1 + \sin A) \)
\[ = \frac{\cos A \times (1 + \sin A)}{(1 - \sin A) \times (1 + \sin A)} = \frac{\cos A(1 + \sin A)}{1 - \sin^2 A} \]
Step 3: Use the identity \( 1 - \sin^2 A = \cos^2 A \)
\[ = \frac{\cos A(1 + \sin A)}{\cos^2 A} \]
Step 4: Cancel common factors
\[ = \frac{1 + \sin A}{\cos A} \]
Step 5: Split the fraction
\[ = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A \]
This equals the right-hand side. Hence proved.
In simple words: We used the same multiplication pattern but with \( (1 + \sin A) \) this time. After using the Pythagorean identity, we split the fraction to get sec and tan.

📝 Teacher's Note: Compare this with Question 35. Show students how the pattern changes when we have \( 1 - \sin A \) vs \( 1 + \sin A \) in the denominator.

🎯 Exam Tip: The multiplication by \( (1 + \sin A) \) is key. Show all steps clearly and write the final trigonometric ratios properly.

Question 37. Prove: \( \frac{\sin A \tan A}{1 - \cos A} = 1 + \sec A \)

Solution:
\( \frac{\sin A \tan A}{1 - \cos A} \)
\( = \frac{\sin A \tan A}{1 - \cos A} \times \frac{1 + \cos A}{1 + \cos A} \)
\( = \frac{\sin A \tan A(1 + \cos A)}{1 - \cos^2 A} \)
\( = \sin A \frac{\sin A}{\cos A} \frac{(1 + \cos A)}{\sin^2 A} \)
\( = \frac{1 + \cos A}{\cos A} \)
\( = \frac{1}{\cos A} + \frac{\cos A}{\cos A} \)
\( = \sec A + 1 \)
In simple words: We multiply top and bottom by (1 + cos A). This makes the bottom become sin²A. Then we simplify to get sec A + 1.

📝 Teacher's Note: Show students the trick of multiplying by the conjugate (1 + cos A). This removes the square root and makes the problem easier.

🎯 Exam Tip: Always multiply by the conjugate when you see (1 - cos A) or (1 + cos A) in the denominator. Write each step clearly.

Question 38. Prove: \( (1 + \cot A - \operatorname{cosec} A)(1 + \tan A + \sec A) = 2 \)

Solution:
\( (1 + \cot A - \operatorname{cosec} A)(1 + \tan A + \sec A) \)
\( = \left(1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A}\right)\left(1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A}\right) \)
\( = \left(\frac{\sin A + \cos A - 1}{\sin A}\right)\left(\frac{\cos A + \sin A + 1}{\cos A}\right) \)
\( = \frac{(\sin A + \cos A - 1)(\sin A + \cos A + 1)}{\sin A \cos A} \)
\( = \frac{(\sin A + \cos A)^2 - (1)^2}{\sin A \cos A} \)
\( = \frac{\sin^2 A + \cos^2 A + 2 \sin A \cos A - 1}{\sin A \cos A} \)
\( = \frac{1 + 2 \sin A \cos A - 1}{\sin A \cos A} \)
\( = \frac{2 \sin A \cos A}{\sin A \cos A} = 2 \)
In simple words: We change everything to sin A and cos A. Then we use the identity (a + b)(a - b) = a² - b². This gives us 2 at the end.

📝 Teacher's Note: Teach students to recognize the pattern (a + b)(a - b) = a² - b². This is the key to solving this problem quickly.

🎯 Exam Tip: When you see expressions like (sin A + cos A - 1) and (sin A + cos A + 1), think of a² - b² formula. Remember sin²A + cos²A = 1.

Question 39. Prove: \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A \)

Solution:
\( \sqrt{\frac{1 + \sin A}{1 - \sin A}} \)
\( = \sqrt{\frac{1 + \sin A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}} \)
\( = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \)
\( = \frac{1 + \sin A}{\cos A} \)
\( = \sec A + \tan A \)
In simple words: We multiply top and bottom by (1 + sin A). This makes the bottom become cos²A. Then we take the square root and split the fraction.

📝 Teacher's Note: Show students that 1 - sin²A = cos²A. This identity is very important for solving square root problems in trigonometry.

🎯 Exam Tip: When you see a square root with 1 ± sin A, multiply by the conjugate. Remember to use 1 - sin²A = cos²A.

Question 40. Prove: \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \operatorname{cosec} A - \cot A \)

Solution:
\( \sqrt{\frac{1 - \cos A}{1 + \cos A}} \)
\( = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}} \)
\( = \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}} \)
\( = \sqrt{\frac{(1 - \cos A)^2}{\sin^2 A}} \)
\( = \frac{1 - \cos A}{\sin A} \)
\( = \operatorname{cosec} A - \cot A \)
In simple words: We multiply top and bottom by (1 - cos A). This makes the bottom become sin²A. Then we take the square root and split the fraction.

📝 Teacher's Note: Remind students that 1 - cos²A = sin²A. This is just another way of writing the basic identity sin²A + cos²A = 1.

🎯 Exam Tip: For square roots with 1 ± cos A, use the conjugate method. Always remember 1 - cos²A = sin²A and write the final answer as cosec A - cot A.

Question 41. Prove: \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} \)

Solution:
\( \sqrt{\frac{1 - \cos A}{1 + \cos A}} \)
\( = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A}} \)
\( = \sqrt{\frac{1 - \cos^2 A}{(1 + \cos A)^2}} \)
\( = \sqrt{\frac{\sin^2 A}{(1 + \cos A)^2}} \)
\( = \frac{\sin A}{1 + \cos A} \)
In simple words: We multiply by (1 + cos A) on top and bottom. This makes the top become sin²A. Then we take the square root.

📝 Teacher's Note: This is a different approach from Question 40. Here we multiply by (1 + cos A) instead of (1 - cos A). Both methods work.

🎯 Exam Tip: You can solve this problem two ways. Choose the method that matches what you need to prove. Always use 1 - cos²A = sin²A.

Question 42. Prove: \( \sqrt{\frac{1 - \sin A}{1 + \sin A}} = \frac{\cos A}{1 + \sin A} \)

Solution:
\( \sqrt{\frac{1 - \sin A}{1 + \sin A}} \)
\( = \sqrt{\frac{1 - \sin A}{1 + \sin A} \times \frac{1 + \sin A}{1 + \sin A}} \)
\( = \sqrt{\frac{1 - \sin^2 A}{(1 + \sin A)^2}} \)
\( = \sqrt{\frac{\cos^2 A}{(1 + \sin A)^2}} \)
\( = \frac{\cos A}{1 + \sin A} \)
In simple words: We multiply by (1 + sin A) on top and bottom. This makes the top become cos²A. Then we take the square root.

📝 Teacher's Note: This is similar to Question 41 but with sin A instead of cos A. The pattern is the same - multiply by the conjugate.

🎯 Exam Tip: Remember 1 - sin²A = cos²A. When you see square roots with trigonometric expressions, always try the conjugate method first.

Question 43. Prove: \( 1 - \frac{\cos^2 A}{1 + \sin A} = \sin A \)

Solution:
\( 1 - \frac{\cos^2 A}{1 + \sin A} \)
\( = \frac{1 + \sin A - \cos^2 A}{1 + \sin A} \)
\( = \frac{\sin A + \sin^2 A}{1 + \sin A} \)
\( = \frac{\sin A(1 + \sin A)}{1 + \sin A} \)
\( = \sin A \)
In simple words: We make a common fraction. Then we use cos²A = 1 - sin²A. This gives us sin A(1 + sin A) on top, which cancels with the bottom.

📝 Teacher's Note: Show students how to substitute cos²A = 1 - sin²A. This substitution is very useful in many trigonometry problems.

🎯 Exam Tip: When you see cos²A in a fraction with sin A, try replacing cos²A with 1 - sin²A. Then factor and cancel common terms.

Question 44. Prove: \( \frac{1}{\sin A + \cos A} + \frac{1}{\sin A - \cos A} = \frac{2 \sin A}{1 - 2 \cos^2 A} \)

Solution:
\( \frac{1}{\sin A + \cos A} + \frac{1}{\sin A - \cos A} \)
\( = \frac{\sin A - \cos A + \sin A + \cos A}{\sin^2 A - \cos^2 A} \)
\( = \frac{2 \sin A}{1 - \cos^2 A - \cos^2 A} = \frac{2 \sin A}{1 - 2 \cos^2 A} \)
In simple words: We add the fractions by making a common denominator. The denominator becomes sin²A - cos²A, which equals 1 - 2cos²A.

📝 Teacher's Note: Teach students that (a + b)(a - b) = a² - b². Also show that sin²A - cos²A = (1 - cos²A) - cos²A = 1 - 2cos²A.

🎯 Exam Tip: When adding fractions with (a + b) and (a - b) in denominators, the common denominator is a² - b². Remember sin²A + cos²A = 1.

Question 45. Prove: \( \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^2 A - 1} \)

Solution:
\( \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} \)
\( = \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{(\sin A + \cos A)(\sin A - \cos A)} \)
\( = \frac{\sin^2 A + \cos^2 A + 2 \sin A \cos A + \sin^2 A + \cos^2 A - 2 \sin A \cos A}{\sin^2 A - \cos^2 A} \)
\( = \frac{2(\sin^2 A + \cos^2 A)}{\sin^2 A - \cos^2 A} \)
\( = \frac{2}{\sin^2 A - \cos^2 A} \) [since sin²A + cos²A = 1]
\( = \frac{2}{\sin^2 A - \cos^2 A} = \frac{2}{\sin^2 A - (1 - \sin^2 A)} \)
\( = \frac{2}{2 \sin^2 A - 1} \)
In simple words: We add the fractions by squaring the numerators. When we expand and simplify, we get 2 on top and 2sin²A - 1 on bottom.

📝 Teacher's Note: Remind students to expand (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b². The middle terms cancel when we add them.

🎯 Exam Tip: When you have fractions like this, add them using a common denominator. Remember cos²A = 1 - sin²A to simplify the final answer.

Question 46.
Prove:
\( \frac{\cot A + \cos \sec A - 1}{\cot A - \cos \sec A + 1} = \frac{1 + \cos A}{\sin A} \)

Solution:
Step 1: Start with the left-hand side (LHS).
\( \frac{\cot A + \cos \sec A - 1}{\cot A - \cos \sec A + 1} \)
Step 2: Use the identity \( \cos^2 A - \cos^2 A = 1 \).
\( = \frac{\cot A + \cos \sec A - (\cos \sec^2 A - \cos^2 A)}{\cot A - \cos \sec A + 1} \)
Step 3: Simplify the numerator using the identity \( \cos \sec^2 A - \cos^2 A = 1 \).
\( = \frac{\cot A + \cos \sec A - [(\cos \sec A - \cos A)(\cos \sec A + \cos A)]}{\cot A - \cos \sec A + 1} \)
Step 4: Factor and simplify.
\( = \frac{\cot A + \cos \sec A [1 - \cos \sec A + \cos A]}{\cot A - \cos \sec A + 1} \)
Step 5: Continue simplification to get \( \cot A + \cos \sec A \).
\( = \frac{\cos A}{\sin A} + \frac{1}{\sin A} \)
Step 6: Combine the fractions.
\( = \frac{1 + \cos A}{\sin A} \)
This equals the right-hand side (RHS).

📝 Teacher's Note: This proof uses the basic trigonometric identities. Students should remember that \( \cot A = \frac{\cos A}{\sin A} \) and \( \sec A = \frac{1}{\cos A} \). Practice these identities well.

🎯 Exam Tip: Write each step clearly. Show all substitutions. The examiner wants to see that you know the basic trigonometric identities.

Question 47.
Prove:
\( \frac{\sin \theta \tan \theta}{1 - \cos \theta} = 1 + \sec \theta \)

Solution:
Step 1: Start with the left-hand side (LHS).
\( \frac{\sin \theta \tan \theta}{1 - \cos \theta} \)
Step 2: Multiply numerator and denominator by \( \frac{1 + \cos \theta}{1 + \cos \theta} \).
\( = \frac{\sin \theta \tan \theta}{1 - \cos \theta} \cdot \frac{1 + \cos \theta}{1 + \cos \theta} \)
Step 3: Simplify using the identity \( 1 - \cos^2 \theta = \sin^2 \theta \).
\( = \frac{\sin \theta \tan \theta (1 + \cos \theta)}{1 - \cos^2 \theta} \)
Step 4: Substitute \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
\( = \frac{\sin \theta \cdot \frac{\sin \theta}{\cos \theta} (1 + \cos \theta)}{\sin^2 \theta} \)
Step 5: Simplify.
\( = \frac{(1 + \cos \theta)}{\cos \theta} \)
Step 6: Split the fraction.
\( = \frac{1}{\cos \theta} + 1 \)
Step 7: Use the identity \( \sec \theta = \frac{1}{\cos \theta} \).
\( = \sec \theta + 1 \)
This equals the right-hand side (RHS).

📝 Teacher's Note: The key trick here is multiplying by the conjugate \( (1 + \cos \theta) \). This creates \( 1 - \cos^2 \theta = \sin^2 \theta \) in the denominator, which cancels nicely.

🎯 Exam Tip: When you see \( 1 - \cos \theta \) in the denominator, try multiplying by \( \frac{1 + \cos \theta}{1 + \cos \theta} \). This is a common exam technique.

Question 48.
Prove:
\( \frac{\cos \theta \cot \theta}{1 + \sin \theta} = \cos \sec \theta - 1 \)

Solution:
Step 1: Start with the left-hand side (LHS).
\( \frac{\cos \theta \cot \theta}{1 + \sin \theta} \)
Step 2: Multiply numerator and denominator by \( \frac{1 - \sin \theta}{1 - \sin \theta} \).
\( = \frac{\cos \theta \cot \theta}{1 + \sin \theta} \cdot \frac{1 - \sin \theta}{1 - \sin \theta} \)
Step 3: Simplify the denominator using \( 1 - \sin^2 \theta = \cos^2 \theta \).
\( = \frac{\cos \theta \cot \theta (1 - \sin \theta)}{1 - \sin^2 \theta} \)
Step 4: Substitute \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and simplify.
\( = \frac{\cos \theta \cdot \frac{\cos \theta}{\sin \theta} (1 - \sin \theta)}{\cos^2 \theta} \)
Step 5: Cancel \( \cos^2 \theta \).
\( = \frac{(1 - \sin \theta)}{\sin \theta} \)
Step 6: Split the fraction.
\( = \frac{1}{\sin \theta} - 1 \)
Step 7: Use the identity \( \csc \theta = \frac{1}{\sin \theta} \).
\( = \csc \theta - 1 \)
This equals the right-hand side (RHS).

📝 Teacher's Note: Similar to the previous problem, we multiply by the conjugate to create \( 1 - \sin^2 \theta \) which equals \( \cos^2 \theta \). This allows for nice cancellation.

🎯 Exam Tip: Remember that \( \csc \theta = \frac{1}{\sin \theta} \). Also, when you see \( 1 + \sin \theta \) in denominator, multiply by \( \frac{1 - \sin \theta}{1 - \sin \theta} \).

Exercise 21 B

Question 1.
Prove that:
(i) \( \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A \)
(ii) \( \frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} + \frac{\cos^3 A - \sin^3 A}{\cos A - \sin A} = 2 \)
(iii) \( \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = \sec A \cos \sec A + 1 \)
(iv) \( \left( \tan A + \frac{1}{\cos A} \right)^2 + \left( \tan A - \frac{1}{\cos A} \right)^2 = 2 \left( \frac{1 + \sin^2 A}{1 - \sin^2 A} \right) \)
(v) \( 2 \sin^2 A + \cos^4 A = 1 + \sin^4 A \)
(vi) \( \frac{\sin A - \sin B}{\cos A + \cos B} + \frac{\cos A - \cos B}{\sin A + \sin B} = 0 \)
(vii) \( (\cos \sec A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A} \)
(viii) \( (1 + \tan A \tan B)^2 + (\tan A - \tan B)^2 = \sec^2 A \sec^2 B \)
(ix) \( \frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = \cos \sec A + \sec A \)

Solution:
(i)
LHS = \( \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} \)
Step 1: Substitute \( \tan A = \frac{\sin A}{\cos A} \) and \( \cot A = \frac{\cos A}{\sin A} \).
\( = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}} \)
Step 2: Simplify the denominators.
\( = \frac{\cos A}{\frac{\cos A - \sin A}{\cos A}} + \frac{\sin A}{\frac{\sin A - \cos A}{\sin A}} \)
Step 3: Simplify by dividing.
\( = \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A} \)
Step 4: Factor out the common term.
\( = \frac{\cos^2 A - \sin^2 A}{(\cos A - \sin A)} \)
Step 5: Use the identity \( \cos^2 A - \sin^2 A = (\cos A - \sin A)(\cos A + \sin A) \).
\( = \sin A + \cos A = \text{RHS} \)
(ii)
\( \frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} + \frac{\cos^3 A - \sin^3 A}{\cos A - \sin A} \)
Step 1: Use the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) and \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).
\( = \frac{(\cos^3 A + \sin^3 A)(\cos A - \sin A) + (\cos^3 A - \sin^3 A)(\cos A + \sin A)}{\cos^2 A - \sin^2 A} \)
Step 2: Expand and simplify.
\( = \frac{\cos^4 A - \cos^3 A \sin A + \sin^3 A \cos A - \sin^4 A + \cos^4 A + \cos^3 A \sin A - \sin^3 A \cos A - \sin^4 A}{\cos^2 A - \sin^2 A} \)
Step 3: Cancel terms and simplify.
\( = \frac{2(\cos^4 A - \sin^4 A)}{\cos^2 A - \sin^2 A} \)
Step 4: Factor using difference of squares.
\( = \frac{2(\cos^2 A + \sin^2 A)(\cos^2 A - \sin^2 A)}{(\cos^2 A - \sin^2 A)} \)
Step 5: Use \( \cos^2 A + \sin^2 A = 1 \).
\( = 2(1) = 2 \)
(iii)
\( \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} \)
Step 1: Substitute the values and simplify step by step.
\( = \frac{\tan A}{1 - \frac{1}{\tan A}} + \frac{\frac{1}{\tan A}}{1 - \tan A} \)
Step 2: Simplify the complex fractions.
\( = \frac{\tan^2 A}{\tan A - 1} + \frac{1}{\tan A(1 - \tan A)} \)
Step 3: Continue algebraic manipulation.
\( = \frac{\tan^3 A - 1}{\tan A(\tan A - 1)} \)
Step 4: Factor and simplify.
\( = \frac{(\tan A - 1)(\tan^2 A + 1 + \tan A)}{\tan A(\tan A - 1)} \)
Step 5: Cancel and use identities to reach the final result.
\( = \frac{\sec^2 A + \tan A}{\tan A} = \frac{1}{\cos^2 A} \cdot \frac{\cos A}{\sin A} + 1 = \frac{1}{\sin A \cos A} + 1 = \sec A \csc A + 1 \)

📝 Teacher's Note: These are complex trigonometric identities. Break each step down carefully. Use the basic identities like \( \sin^2 A + \cos^2 A = 1 \) and \( \tan A = \frac{\sin A}{\cos A} \) repeatedly.

🎯 Exam Tip: Show every algebraic step clearly. The examiner wants to see your working, not just the final answer. Practice factoring expressions using \( a^3 + b^3 \) and \( a^3 - b^3 \) formulas.

Question 2. If \( x \cos A + y \sin A = m \) and \( x \sin A - y \cos A = n \), then prove that \( x^2 + y^2 = m^2 + n^2 \).
Answer:
\( m^2 + n^2 \)
\( = (x \cos A + y \sin A)^2 + (x \sin A - y \cos A)^2 \)
\( = x^2 \cos^2 A + y^2 \sin^2 A + 2xy \sin A \cos A \)
\( + x^2 \sin^2 A + y^2 \cos^2 A - 2xy \sin A \cos A \)
\( = x^2 (\cos^2 A + \sin^2 A) + y^2 (\cos^2 A + \sin^2 A) \)
\( = x^2 + y^2 \)
Hence, \( x^2 + y^2 = m^2 + n^2 \).
In simple words: We square both given equations and add them. The cross terms cancel out and we use the identity \( \sin^2 A + \cos^2 A = 1 \).

📝 Teacher's Note: This is a very common type of proof question. Show students how to expand the squares carefully and point out how the mixed terms cancel out perfectly.

🎯 Exam Tip: Always expand both squares completely first. Write "\( \sin^2 A + \cos^2 A = 1 \)" clearly - this is the key step that gets you marks.

Question 3. If \( m = a \sec A + b \tan A \) and \( n = a \tan A + b \sec A \), prove that \( m^2 - n^2 = a^2 - b^2 \)
Answer:
Given,
\( m = a \sec A + b \tan A \) and \( n = a \tan A + b \sec A \)
\( m^2 - n^2 = (a \sec A + b \tan A)^2 - (a \tan A + b \sec A)^2 \)
\( = a^2 \sec^2 A + b^2 \tan^2 A + 2ab \sec A \tan A \)
\( - (a^2 \tan^2 A + b^2 \sec^2 A + 2ab \sec A \tan A) \)
\( = \sec^2 A(a^2 - b^2) + \tan^2 A(b^2 - a^2) \)
\( = (a^2 - b^2)[\sec^2 A - \tan^2 A] \)
\( = (a^2 - b^2) \) [Since \( \sec^2 A - \tan^2 A = 1 \)]
Hence, \( m^2 - n^2 = a^2 - b^2 \)
In simple words: We expand the squares and group similar terms. Then we use the identity \( \sec^2 A - \tan^2 A = 1 \) to simplify.

📝 Teacher's Note: Emphasize the identity \( \sec^2 A - \tan^2 A = 1 \). Students often forget this key trigonometric identity. Practice it separately first.

🎯 Exam Tip: Write the identity "\( \sec^2 A - \tan^2 A = 1 \)" clearly at the end. This shows you know the key step and gets full marks.

Question 4. If \( x = r \sin A \cos B \), \( y = r \sin A \sin B \) and \( z = r \cos A \), prove that \( x^2 + y^2 + z^2 = r^2 \)
Answer:
LHS = \( (r \sin A \cos B)^2 + (r \sin A \sin B)^2 + (r \cos A)^2 \)
\( = r^2 \sin^2 A \cos^2 B + r^2 \sin^2 A \sin^2 B + r^2 \cos^2 A \)
\( = r^2 \sin^2 A(\cos^2 B + \sin^2 B) + r^2 \cos^2 A \)
\( = r^2(\sin^2 A + \cos^2 A) = r^2 \) = RHS
In simple words: We square all three terms, factor out common parts, and use the identity \( \sin^2 + \cos^2 = 1 \) twice.

📝 Teacher's Note: This is like finding the distance formula in 3D space. Show students how factoring makes the calculation much easier.

🎯 Exam Tip: Factor out \( r^2 \) first, then use \( \sin^2 + \cos^2 = 1 \). Write "= RHS" at the end to show completion.

Question 5. If \( \sin A + \cos A = m \) and \( \sec A + \cosec A = n \), prove that \( n(m^2 - 1) = 2m \).
Answer:
Given:
\( \sin A + \cos A = m \)
and
\( \sec A + \cosec A = n \)
Consider L.H.S = \( n(m^2 - 1) \)
\( = (\sec A + \cosec A)[(\sin A + \cos A)^2 - 1] \)
\( = \left(\frac{1}{\cos A} + \frac{1}{\sin A}\right)[\sin^2 A + \cos^2 A + 2 \sin A \cos A - 1] \)
\( = \left(\frac{\cos A + \sin A}{\sin A \cos A}\right)(1 + 2 \sin A \cos A - 1) \)
\( = \frac{(\cos A + \sin A)}{\sin A \cos A}(2 \sin A \cos A) \)
\( = 2(\sin A + \cos A) \)
\( = 2m = \) R.H.S.
In simple words: We substitute the given values, expand the square, and simplify step by step using basic algebra.

📝 Teacher's Note: Show students how to convert sec and cosec to fractions first. This makes the algebra much clearer to follow.

🎯 Exam Tip: Write "Given:" clearly at the start. Show each substitution step and write "= R.H.S." at the end for full marks.

Question 6. If \( x = r \cos A \cos B \), \( y = r \cos A \sin B \) and \( z = r \sin A \), prove that \( x^2 + y^2 + z^2 = r^2 \)
Answer:
LHS = \( (r \cos A \cos B)^2 + (r \cos A \sin B)^2 + (r \sin A)^2 \)
\( = r^2 \cos^2 A \cos^2 B + r^2 \cos^2 A \sin^2 B + r^2 \sin^2 A \)
\( = r^2 \cos^2 A(\cos^2 B + \sin^2 B) + r^2 \sin^2 A \)
\( = r^2(\cos^2 A + \sin^2 A) = r^2 \) = RHS
In simple words: Similar to Question 4, we square all terms, factor out common parts, and use \( \sin^2 + \cos^2 = 1 \).

📝 Teacher's Note: Point out the similarity with Question 4. Both use the same technique - factoring and the Pythagorean identity.

🎯 Exam Tip: Factor systematically and use the identity clearly. This type of question appears frequently in exams.

Question 7. If \( \frac{\cos A}{\cos B} = m \) and \( \frac{\cos A}{\sin B} = n \), show that \( (m^2 + n^2) \cos^2 B = n^2 \).
Answer:
LHS = \( (m^2 + n^2) \cos^2 B \)
\( = \left(\frac{\cos^2 A}{\cos^2 B} + \frac{\cos^2 A}{\sin^2 B}\right) \cos^2 B \)
\( = \left(\frac{\cos^2 A \sin^2 B + \cos^2 A \cos^2 B}{\cos^2 B \sin^2 B}\right) \cos^2 B \)
\( = \left(\frac{\cos^2 A \sin^2 B + \cos^2 A \cos^2 B}{\sin^2 B}\right) \)
\( = \frac{\cos^2 A(\sin^2 B + \cos^2 B)}{\sin^2 B} \)
\( = \frac{\cos^2 A}{\sin^2 B} \)
\( = n^2 \)
Hence, \( (m^2 + n^2) \cos^2 B = n^2 \).
In simple words: We substitute the given ratios, combine fractions, and use \( \sin^2 B + \cos^2 B = 1 \) to simplify.

📝 Teacher's Note: Show students how to work with algebraic fractions step by step. The key is careful substitution and fraction manipulation.

🎯 Exam Tip: Substitute the given values for m and n carefully. Show all fraction operations clearly for full marks.

Exercise 21 C

Question 1. Without using trigonometric tables, show that:
(i) \( \tan 10° \tan 15° \tan 75° \tan 80° = 1 \)
(ii) \( \sin 42° \sec 48° + \cos 42° \cos ec 48° = 2 \)
(iii) \( \frac{\sin 26°}{\sec 64°} + \frac{\cos 26°}{\cos ec 64°} = 1 \)
Answer:
(i)
\( \tan 10° \tan 15° \tan 75° \tan 80° \)
\( = \tan(90° - 80°) \tan(90° - 75°) \tan 75° \tan 80° \)
\( = \cot 80° \cot 75° \tan 75° \tan 80° \)
\( = 1 \) [As \( \tan θ \cot θ = 1 \)]
(ii)
\( \sin 42° \sec 48° + \cos 42° \cos ec 48° = 2 \)
Consider \( \sin 42° \sec 48° + \cos 42° \cos ec 48° \)
\( ⇒ \sin 42° \sec(90° - 42°) + \cos 42° \cos ec(90° - 42°) \)
\( ⇒ \sin 42° \cos ec 42° + \cos 42° \sec 42° \)
\( ⇒ \sin 42° \frac{1}{\sin 42°} + \cos 42° \frac{1}{\cos 42°} \)
\( ⇒ 1 + 1 = 2 \)
(iii)
\( \frac{\sin 26°}{\sec 64°} + \frac{\cos 26°}{\cos ec 64°} \)
\( = \frac{\sin 26°}{\sec(90° - 26°)} + \frac{\cos 26°}{\cos ec(90° - 26°)} \)
\( = \frac{\sin 26°}{\cos ec 26°} + \frac{\cos 26°}{\sec 26°} \)
\( = \sin^2 26° + \cos^2 26° \)
\( = 1 \)
In simple words: We use complementary angle relationships like \( \tan(90° - θ) = \cot θ \) and \( \sin^2 θ + \cos^2 θ = 1 \).

📝 Teacher's Note: Teach the complementary angle pairs first: sin-cos, tan-cot, sec-cosec. Students need to memorize these relationships.

🎯 Exam Tip: Always look for complementary angles (angles that add to 90°). Write the complementary relationships clearly in your working.

Question 2. Express each of the following in terms of angles between 0° and 45°:
(i) \( \sin 59° + \tan 63° \)
(ii) \( \cos ec 68° + \cot 72° \)
(iii) \( \cos 74° + \sec 67° \)
Answer:
(i) \( \sin 59° + \tan 63° \)
\( = \sin(90 - 31)° + \tan(90 - 27)° \)
\( = \cos 31° + \cot 27° \)
(ii) \( \cos ec 68° + \cot 72° \)
\( = \cos ec (90 - 22)° + \cot(90 - 18)° \)
\( = \sec 22° + \tan 18° \)
(iii) \( \cos 74° + \sec 67° \)
\( = \cos(90 - 16)° + \sec(90 - 23)° \)
\( = \sin 16° + \cos ec 23° \)
In simple words: We use the fact that trigonometric functions of complementary angles are related. For angles above 45°, we subtract from 90° and change the function.

📝 Teacher's Note: Show students the pattern: sin becomes cos, tan becomes cot, sec becomes cosec when we use complementary angles.

🎯 Exam Tip: For any angle above 45°, write it as (90° - something) first. Then use complementary angle relationships.

Question 3. Show that:
(i) \( \frac{\sin A}{\sin(90° - A)} + \frac{\cos A}{\cos(90° - A)} = \sec A \cos ec A \)
(ii) \( \sin A \cos A - \frac{\sin A \cos(90° - A) \cos A}{\sec(90° - A)} - \frac{\cos A \sin(90° - A) \sin A}{\cos ec(90° - A)} = 0 \)
Answer:
(i) \( \frac{\sin A}{\sin(90° - A)} + \frac{\cos A}{\cos(90° - A)} \)
\( = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \)
\( = \frac{\sin^2 A + \cos^2 A}{\cos A \sin A} \)
\( = \frac{1}{\cos A \sin A} \)
\( = \sec A \cos ec A \)
(ii) \( \sin A \cos A - \frac{\sin A \cos(90° - A) \cos A}{\sec(90° - A)} - \frac{\cos A \sin(90° - A) \sin A}{\cos ec(90° - A)} \)
\( = \sin A \cos A - \frac{\sin A \sin A \cos A}{\cos ec A} - \frac{\cos A \cos A \sin A}{\sec A} \)
\( = \sin A \cos A - \sin^2 A \cos A - \cos^3 A \sin A \)
\( = \sin A \cos A - \sin A \cos A(\sin^2 A + \cos^2 A) \)
\( = \sin A \cos A - \sin A \cos A(1) \)
\( = 0 \)
In simple words: We substitute complementary angle relationships and simplify using algebraic manipulation and the Pythagorean identity.

📝 Teacher's Note: Break down part (ii) step by step. Students often get confused with the multiple substitutions. Do each substitution separately.

🎯 Exam Tip: For part (i), remember to get a common denominator. For part (ii), substitute complementary angles first before simplifying.

Question 4. For triangle ABC, show that:
(i) \( \sin\left(\frac{A + B}{2}\right) = \cos\frac{C}{2} \)
(ii) \( \tan\left(\frac{B + C}{2}\right) = \cot\frac{A}{2} \)
Answer:
In triangle ABC, \( A + B + C = 180° \)
(i) \( A + B = 180° - C \)
\( ⇒ \frac{A + B}{2} = \frac{180° - C}{2} = 90° - \frac{C}{2} \)
\( ⇒ \sin\left(\frac{A + B}{2}\right) = \sin\left(90° - \frac{C}{2}\right) = \cos\frac{C}{2} \)
(ii) \( B + C = 180° - A \)
\( ⇒ \frac{B + C}{2} = \frac{180° - A}{2} = 90° - \frac{A}{2} \)
\( ⇒ \tan\left(\frac{B + C}{2}\right) = \tan\left(90° - \frac{A}{2}\right) = \cot\frac{A}{2} \)
In simple words: In any triangle, the sum of all angles is 180°. We use this fact and complementary angle relationships.

📝 Teacher's Note: Start with the angle sum property of triangles. This is the key fact students need to remember for all triangle problems.

🎯 Exam Tip: Always write "In triangle ABC, A + B + C = 180°" at the start. This shows you know the basic property and gets you marks.

Solution:

(i) We know that for a triangle \( \triangle ABC \)
\( \angle A + \angle B + \angle C = 180° \)
\( \frac{\angle B + \angle A}{2} = 90° - \frac{\angle C}{2} \)
\( \sin\left(\frac{A + B}{2}\right) = \sin\left(90° - \frac{C}{2}\right) \)
\( = \cos\left(\frac{C}{2}\right) \)

(ii) We know that for a triangle \( \triangle ABC \)
\( \angle A + \angle B + \angle C = 180° \)
\( \angle B + \angle C = 180° - \angle A \)
\( \frac{\angle B + \angle C}{2} = 90° - \frac{\angle A}{2} \)
\( \tan\left(\frac{B + C}{2}\right) = \tan\left(90° - \frac{A}{2}\right) \)
\( = \cot\left(\frac{A}{2}\right) \)

📝 Teacher's Note: In any triangle, the sum of all three angles is always 180°. Use this fact to convert half-angle formulas. Students often forget to divide by 2 when working with half angles.

🎯 Exam Tip: Always start by writing the angle sum property. Then rearrange to get the angle you need. Show each step clearly to get full marks.

Question 5. Evaluate:
(i) \( 3\frac{\sin 72°}{\cos 18°} - \frac{\sec 32°}{\cos ec 58°} \)
(ii) \( 3\cos 80° \cos ec 10° + 2\cos 59° \cos ec 31° \)
(iii) \( \frac{\sin 80°}{\cos 10°} + \sin 59° \sec 31° \)
(iv) \( \tan(55° - A) - \cot(35° + A) \)
(v) \( \cos ec(65° + A) - \sec(25° - A) \)
(vi) \( 2\frac{\tan 57°}{\cot 33°} - \frac{\cot 70°}{\tan 20°} - \sqrt{2}\cos 45° \)
(vii) \( \frac{\cot^2 41°}{\tan^2 49°} - 2\frac{\sin^2 75°}{\cos^2 15°} \)
(viii) \( \frac{\cos 70°}{\sin 20°} + \frac{\cos 59°}{\sin 31°} - 8\sin^2 30° \)
(ix) \( 14\sin 30° + 6\cos 60° - 5\tan 45° \)

Solution:

(i) \( 3\frac{\sin 72°}{\cos 18°} - \frac{\sec 32°}{\cos ec 58°} \)
\( = 3\frac{\sin(90° - 18°)}{\cos 18°} - \frac{\sec(90° - 58°)}{\cos ec 58°} \)
\( = 3\frac{\cos 18°}{\cos 18°} - \frac{\cos ec 58°}{\cos ec 58°} = 3 - 1 = 2 \)

(ii) \( 3\cos 80° \cos ec 10° + 2\cos 59° \cos ec 31° \)
\( = 3\cos(90° - 10°)\cos ec 10° + 2\cos(90° - 31°)\cos ec 31° \)
\( = 3\sin 10° \cos ec 10° + 2\sin 31° \cos ec 31° \)
\( = 3 + 2 = 5 \)

(iii) \( \frac{\sin 80°}{\cos 10°} + \sin 59° \sec 31° \)
\( = \frac{\sin(90° - 10°)}{\cos 10°} + \sin(90° - 31°)\sec 31° \)
\( = \frac{\cos 10°}{\cos 10°} + \frac{\cos 31°}{\cos 31°} \)
\( = 1 + 1 = 2 \)

(iv) \( \tan(55° - A) - \cot(35° + A) \)
\( = \tan[90° - (35° + A)] - \cot(35° + A) \)
\( = \cot(35° + A) - \cot(35° + A) \)
\( = 0 \)

(v) \( \cos ec(65° + A) - \sec(25° - A) \)
\( = \cos ec[90° - (25° - A)] - \sec(25° - A) \)
\( = \sec(25° - A) - \sec(25° - A) \)
\( = 0 \)

(vi) \( 2\frac{\tan 57°}{\cot 33°} - \frac{\cot 70°}{\tan 20°} - \sqrt{2}\cos 45° \)
\( = 2\frac{\tan(90° - 33°)}{\cot 33°} - \frac{\cot(90° - 20°)}{\tan 20°} - \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) \)
\( = 2\frac{\cot 33°}{\cot 33°} - \frac{\tan 20°}{\tan 20°} - 1 \)
\( = 2 - 1 - 1 \)
\( = 0 \)

(vii) \( \frac{\cot^2 41°}{\tan^2 49°} - 2\frac{\sin^2 75°}{\cos^2 15°} \)
\( = \frac{[\cot(90° - 49°)]^2}{\tan^2 49°} - 2\frac{[\sin(90° - 15°)]^2}{\cos^2 15°} \)
\( = \frac{\tan^2 49°}{\tan^2 49°} - 2\frac{\cos^2 15°}{\cos^2 15°} \)
\( = 1 - 2 = -1 \)

(viii) \( \frac{\cos 70°}{\sin 20°} + \frac{\cos 59°}{\sin 31°} - 8\sin^2 30° \)
\( = \frac{\cos(90° - 20°)}{\sin 20°} + \frac{\cos(90° - 31°)}{\sin 31°} - 8\left(\frac{1}{2}\right)^2 \)
\( = \frac{\sin 20°}{\sin 20°} + \frac{\sin 31°}{\sin 31°} - 2 \)
\( = 1 + 1 - 2 = 0 \)

(ix) \( 14\sin 30° + 6\cos 60° - 5\tan 45° \)
\( = 14\left(\frac{1}{2}\right) + 6\left(\frac{1}{2}\right) - 5(1) \)
\( = 7 + 3 - 5 = 5 \)

In simple words: These problems use complementary angle rules. When two angles add up to 90°, their trigonometric functions are related. For example, sin 72° = cos 18° because 72° + 18° = 90°.

📝 Teacher's Note: Teach students that sin θ = cos(90° - θ) and tan θ = cot(90° - θ). Make them memorize these complementary angle pairs. Use a right triangle to show why this works.

🎯 Exam Tip: Always look for complementary angles first. Convert using the complementary rules, then simplify. Write each step clearly - examiners give marks for working, not just the final answer.

Question 6. A triangle ABC is right angled at B; find the value of \( \frac{\sec A \cos ec C - \tan A \cot C}{\sin B} \)

Solution:
Since, ABC is a right angled triangle, right angled at B.
So, A + C = 90°
\( \frac{\sec A \cos ec C - \tan A \cot C}{\sin B} \)
\( = \frac{\sec(90° - C) \cos ec C - \tan(90° - C) \cot C}{\sin 90°} \)
\( = \frac{\cos ec C \cos ec C - \cot C \cot C}{1} \)
\( = 1 \) [∵ \( \cos ec^2 θ - \cot^2 θ = 1 \)]

In simple words: In a right triangle, if one angle is 90°, the other two angles add up to 90°. We used this fact and the trigonometric identity \( \cos ec^2 θ - \cot^2 θ = 1 \) to solve this.

📝 Teacher's Note: Draw a right triangle and label the angles. Show that A + C = 90° when B = 90°. Students must memorize the identity \( \cos ec^2 θ - \cot^2 θ = 1 \).

🎯 Exam Tip: For right triangles, always write A + C = 90° first. Then use complementary angle rules. The answer will often be a simple number like 0 or 1.

Question 7. Find (in each case, given below) the value of x if:
(i) \( \sin x = \sin 60° \cos 30° - \cos 60° \sin 30° \)
(ii) \( \sin x = \sin 60° \cos 30° + \cos 60° \sin 30° \)
(iii) \( \cos x = \cos 60° \cos 30° - \sin 60° \sin 30° \)
(iv) \( \tan x = \frac{\tan 60° - \tan 30°}{1 + \tan 60° \tan 30°} \)
(v) \( \sin 2x = 2\sin 45° \cos 45° \)
(vi) \( \sin 3x = 2\sin 30° \cos 30° \)
(vii) \( \cos(2x - 6°) = \cos^2 30° - \cos^2 60° \)

Solution:

(i) \( \sin x = \sin 60° \cos 30° - \cos 60° \sin 30° \)
\( \sin x = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \)
\( \sin x = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = \sin 30° \)
Hence, \( x = 30° \)

(ii) \( \sin x = \sin 60° \cos 30° + \cos 60° \sin 30° \)
\( \sin x = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \)
\( \sin x = \frac{3}{4} + \frac{1}{4} = 1 = \sin 90° \)
Hence, \( x = 90° \)

(iii) \( \cos x = \cos 60° \cos 30° - \sin 60° \sin 30° \)
\( \cos x = \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) \)
\( \cos x = 0 = \cos 90° \)
Hence, \( x = 90° \)

(iv) \( \tan x = \frac{\tan 60° - \tan 30°}{1 + \tan 60° \tan 30°} \)
\( \tan x = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}}} \)
\( \tan x = \frac{\frac{3-1}{\sqrt{3}}}{1 + 1} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} = \tan 30° \)
Hence, \( x = 30° \)

(v) \( \sin 2x = 2\sin 45° \cos 45° \)
\( \sin 2x = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{1}{2} = 1 = \sin 90° \)
\( 2x = 90° \)
\( x = 45° \)

(vi) \( \sin 3x = 2\sin 30° \cos 30° \)
\( \sin 3x = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} = \sin 60° \)
\( 3x = 60° \)
\( x = 20° \)

(vii) \( \cos(2x - 6°) = \cos^2 30° - \cos^2 60° \)
\( \cos(2x - 6°) = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \)
\( \cos(2x - 6°) = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = \cos 60° \)
\( 2x - 6° = 60° \)
\( 2x = 66° \)
\( x = 33° \)

In simple words: These problems use addition and subtraction formulas for trigonometry. We substitute the standard values like sin 30° = 1/2, cos 60° = 1/2, etc., then simplify to find x.

📝 Teacher's Note: Make students memorize the standard values: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2. Also teach the sum formulas: sin(A ± B) = sin A cos B ± cos A sin B.

🎯 Exam Tip: Always substitute standard values first, then simplify step by step. Show all working clearly. Remember to check if your final angle makes sense (usually between 0° and 90°).

Question 8. (continuation)

(v) \( \sin 2x = 2 \sin 45° \cos 45° \)
\( \sin 2x = 2 \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) \)
\( \sin 2x = 1 = \sin 90° \)
\( 2x = 90° \)
Hence, \( x = 45° \)

(vi) \( \sin 3x = 2 \sin 30° \cos 30° \)
\( \sin 3x = 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \)
\( \sin 3x = \frac{\sqrt{3}}{2} = \sin 60° \)
\( 3x = 60° \)
Hence, \( x = 20° \)

(vii) \( \cos(2x - 6°) = \cos^2 30° - \cos^2 60° \)
\( \cos(2x - 6) = \cos^2(90° - 60°) - \cos^2 60° \)
\( \cos(2x - 6) = \sin^2 60° - \cos^2 60° \)
\( \cos(2x - 6) = 1 - 2 \cos^2 60° = 1 - 2\left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \cos(2x - 6) = \frac{1}{2} \)
\( \cos(2x - 6) = \cos 60° \)
\( (2x - 6) = 60° \)
\( 2x = 66° \)
Hence, \( x = 33° \)

Question 8. In each case, given below, find the value of angle A, where \( 0° ≤ A ≤ 90° \).
(i) \( \sin(90° - 3A) \cdot \cosec 42° = 1 \)
(ii) \( \cos(90° - A) \cdot \sec 77° = 1 \)

Solution:
(i) \( \sin(90° - 3A) \cdot \cosec 42° = 1 \)
\( \cos 3A \cdot \frac{1}{\sin 42°} = 1 \)
\( \cos 3A = \sin 42° = \sin(90° - 48°) = \cos 48° \)
\( 3A = 48° \)
\( A = 16° \)
(ii) \( \cos(90° - A) \cdot \sec 77° = 1 \)
\( \cos(90° - A) \cdot \sec 77° = 1 \)
\( \sin A \cdot \frac{1}{\cos 77°} = 1 \)
\( \sin A = \cos 77° = \cos(90° - 13°) = \sin 13° \)
\( A = 13° \)

📝 Teacher's Note: Use the complementary angle rules: \( \sin(90° - x) = \cos x \) and \( \cos(90° - x) = \sin x \). Also remember that \( \cosec θ = \frac{1}{\sin θ} \) and \( \sec θ = \frac{1}{\cos θ} \).

🎯 Exam Tip: First convert all functions to sine or cosine using complementary angle rules. Then compare angles directly to find the value.

Question 9. Prove that:
(i) \( \frac{\cos(90° - θ)\cos θ}{\cot θ} = 1 - \cos^2 θ \)
(ii) \( \frac{\sin θ \sin(90° - θ)}{\cot(90° - θ)} = 1 - \sin^2 θ \)

Solution:
(i)
LHS = \( \frac{\cos(90° - θ)\cos θ}{\cot θ} = \frac{\sin θ \cos θ}{\cos θ} \cdot \frac{\sin θ}{\cos θ} = \sin^2 θ = 1 - \cos^2 θ \)
(ii)
LHS = \( \frac{\sin θ \sin(90° - θ)}{\cot(90° - θ)} = \frac{\sin θ \cos θ}{\tan θ} = \frac{\sin θ \cos θ}{\frac{\sin θ}{\cos θ}} = \cos^2 θ = 1 - \sin^2 θ \)

📝 Teacher's Note: Remember these key identities: \( \sin^2 θ + \cos^2 θ = 1 \), \( \cot θ = \frac{\cos θ}{\sin θ} \), and \( \sin(90° - θ) = \cos θ \). Use them step by step.

🎯 Exam Tip: Always write "LHS" and "RHS" clearly. Show each step. Use the identity \( \sin^2 θ + \cos^2 θ = 1 \) to get the final answer.

Question 10. Evaluate: \( \frac{\sin 35° \cos 55° + \cos 35° \sin 55°}{\cos \sec^2 10° - \tan^2 80°} \)

Solution:
\( \frac{\sin 35° \cos 55° + \cos 35° \sin 55°}{\cos \sec^2 10° - \tan^2 80°} \)
= \( \frac{\sin 35° \cos (90° - 35°) + \cos 35° \sin (90° - 35°)}{\cos \sec^2 (90° - 80°) - \tan^2 80°} \)
= \( \frac{\sin 35° \sin 35° + \cos 35° \cos 35°}{\sec^2 80° - \tan^2 80°} \)
= \( \frac{\sin^2 35° + \cos^2 35°}{\sec^2 80° - \tan^2 80°} = \frac{1}{1} = 1 \)

📝 Teacher's Note: Use the complementary angle rules: \( \cos(90° - θ) = \sin θ \). Also use the identity \( \sec^2 θ - \tan^2 θ = 1 \) and \( \sin^2 θ + \cos^2 θ = 1 \).

🎯 Exam Tip: Convert all angles to complementary forms first. Then use basic trigonometric identities to simplify. The answer will be 1.

Question 11. Evaluate \( \sin^2 34° + \sin^2 56° + 2\tan 18° \tan 72° - \cot^2 30° \)

Solution:
\( \sin^2 34° + \sin^2 56° + 2\tan 18° \tan 72° - \cot^2 30° \)
= \( \sin^2 34° + \sin^2(90° - 34°) + 2\tan 18° \tan(90° - 72°) - \cot^2 30° \)
= \( \sin^2 34° + \cos^2 34° + 2\tan 18° \cot 18° - \cot^2 30° \)
= \( (\sin^2 34° + \cos^2 34°) + 2\tan 18° × \frac{1}{\tan 18°} - \cot^2 30° \)
= \( 1 + 2 × 1 - (\sqrt{3})^2 \)
= \( 1 + 2 - 3 \)
= \( 3 - 3 \)
= 0

📝 Teacher's Note: Use complementary angles: \( \sin(90° - θ) = \cos θ \) and \( \tan(90° - θ) = \cot θ \). Also remember \( \tan θ × \cot θ = 1 \) and \( \cot 30° = \sqrt{3} \).

🎯 Exam Tip: Convert to complementary angles first. Use the identity \( \sin^2 θ + \cos^2 θ = 1 \) and \( \tan θ \cot θ = 1 \). The final answer is 0.

Question 12. Without using trigonometrical tables, evaluate: \( \cosec^2 57° - \tan^2 33° + \cos 44° \cosec 46° - \sqrt{2}\cos 45° - \tan^2 60° \)

Solution:
\( \cosec^2 57° - \tan^2 33° + \cos 44° \cosec 46° - \sqrt{2} \cos 45° - \tan^2 60° \)
= \( \cosec^2(90° - 33°) - \tan^2 33° + \cos 44° \cosec (90° - 44°) - \sqrt{2} \cos 45° - \tan^2 60° \)
= \( \sec^2 33° - \tan^2 33° + \cos 44° \sec 44° - \sqrt{2} \cos 45° - \tan^2 60 \)
= \( 1 + 1 - \sqrt{2} \cos 45° - \tan^2 60 \)
= \( 1 + 1 - \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) - (\sqrt{3})^2 \)
= \( 2 - 1 - 3 \)
= -2

📝 Teacher's Note: Use complementary angle identities and basic identities like \( \sec^2 θ - \tan^2 θ = 1 \). Remember \( \cos 45° = \frac{1}{\sqrt{2}} \) and \( \tan 60° = \sqrt{3} \).

🎯 Exam Tip: Convert everything to complementary angles. Use the identity \( \sec^2 θ - \tan^2 θ = 1 \). Calculate standard angle values carefully to get -2.

Exercise 21 D

Question 1. Use tables to find sine of:
(i) 21°
(ii) 34° 42'
(iii) 47° 32'
(iv) 62° 57'
(v) 10° 20' + 20° 45'

Solution:
(i) sin 21° = 0.3584
(ii) sin 34° 42' = 0.5693
(iii) sin 47° 32' = sin (47° 30' + 2') = 0.7373 + 0.0004 = 0.7377
(iv) sin 62° 57' = sin (62° 54' + 3') = 0.8902 + 0.0004 = 0.8906
(v) sin (10° 20' + 20° 45') = sin 30°65' = sin 31°5' = 0.5150 + 0.0012 = 0.5162

📝 Teacher's Note: When using tables, find the nearest tabulated value first. Then use proportional parts to add the difference for extra minutes. Show students how to read trigonometric tables step by step.

🎯 Exam Tip: Always convert minutes to the nearest tabulated value plus extra minutes. Use proportional parts table to find the correction. Write answers to 4 decimal places.

Question 2. Use tables to find cosine of:
(i) 2° 4'
(ii) 8° 12'
(iii) 26° 32'
(iv) 65° 41'
(v) 9° 23' + 15° 54'

Solution:
(i) cos 2° 4' = 0.9994 - 0.0001 = 0.9993
(ii) cos 8° 12' = cos 0.9898
(iii) cos 26° 32' = cos (26° 30' + 2') = 0.8949 - 0.0003 = 0.8946
(iv) cos 65° 41' = cos (65° 36' + 5') = 0.4131 -0.0013 = 0.4118
(v) cos (9° 23' + 15° 54') = cos 24° 77' = cos 25° 17' = cos (25° 12' + 5') = 0.9048 - 0.0006 = 0.9042

📝 Teacher's Note: For cosine, the proportional parts are usually subtracted (not added) because cosine decreases as angle increases. Teach students this important difference from sine.

🎯 Exam Tip: Remember that for cosine, you subtract the proportional part correction. For sine, you add it. This is because cosine decreases as angle increases.

Question 3. Use trigonometrical tables to find tangent of:
(i) 37°
(ii) 42° 18'
(iii) 17° 27'

Solution:
(i) tan 37° = 0.7536
(ii) tan 42° 18' = 0.9099
(iii) tan 17° 27' = tan (17° 24' + 3') = 0.3134 + 0.0010 = 0.3144

📝 Teacher's Note: For tangent tables, use proportional parts by adding the correction for extra minutes. Tangent increases as angle increases, so we add the correction like in sine.

🎯 Exam Tip: For tangent, add the proportional part correction just like in sine. Read the tangent table carefully and use proportional parts for minutes not directly given.

Question 4. Use tables to find the acute angle θ, if the value of sin θ is:
(i) 0.4848
(ii) 0.3827
(iii) 0.6525

Solution:
(i) From the tables, it is clear that sin 29° = 0.4848
Hence, θ = 29°
(ii) From the tables, it is clear that sin 22° 30' = 0.3827
Hence, θ = 22° 30'
(iii) From the tables, it is clear that sin 40° 42' = 0.6521
sin θ - sin 40° 42' = 0.6525 - 0.6521 = 0.0004
From the tables, diff of 2' = 0.0004
Hence, θ = 40° 42' + 2' = 40° 44'

📝 Teacher's Note: To find angle from sine value, look up the sine table in reverse. If exact value is not found, find the closest smaller value and use proportional parts to find extra minutes.

🎯 Exam Tip: When finding angle from sine value, look for the closest value in tables first. Then find the difference and use proportional parts to get extra minutes.

Question 5. Use tables to find the acute angle θ, if the value of cos θ is:
(i) 0.9848
(ii) 0.9574
(iii) 0.6885

📝 Teacher's Note: For finding angles from cosine values, use the cosine table in reverse. Remember that cosine decreases as angle increases, so be careful with proportional parts.

🎯 Exam Tip: Look up cosine values in the table. If exact value is not found, find the difference and use proportional parts. Remember cosine decreases as angle increases.

Question 6. Use tables to find the acute angle θ, if the value of tan q is:
(i) 0.2419
(ii) 0.4741
(iii) 0.7391

Answer:
(i) From the tables, it is clear that cos 10° = 0.9848
Hence, θ = 10°
(ii) From the tables, it is clear that cos 16° 48' = 0.9573
cos θ - cos 16° 48' = 0.9574 - 0.9573 = 0.0001
From the tables, diff of 1' = 0.0001
Hence, θ = 16° 48' - 1' = 16° 47'
(iii) From the tables, it is clear that cos 46° 30' = 0.6884
cos q - cos 46° 30' = 0.6885 - 0.6884 = 0.0001
From the tables, diff of 1' = 0.0002
Hence, θ = 46° 30' - 1' = 46° 29'

In simple words: We look up the tan values in tables. When we find the closest value, we get the angle. Sometimes we need to adjust by a few minutes using the difference method.

📝 Teacher's Note: Show students how to use trigonometric tables. Explain that when exact values are not in the table, we use the difference method to find precise angles.

🎯 Exam Tip: Always check if your answer needs adjustment using the difference method. Write the angle clearly with degrees and minutes.

Question 6. Use tables to find the acute angle θ, if the value of tan q is:
(i) 0.2419
(ii) 0.4741
(iii) 0.7391

Answer:
(i) From the tables, it is clear that tan 13° 36' = 0.2419
Hence, θ = 13° 36'
(ii) From the tables, it is clear that tan 25° 18' = 0.4727
tan θ - tan 25° 18' = 0.4741 - 0.4727 = 0.0014
From the tables, diff of 4' = 0.0014
Hence, θ = 25° 18' + 4' = 25° 22'
(iii) From the tables, it is clear that tan 36° 24' = 0.7373
tan θ - tan 36° 24' = 0.7391 - 0.7373 = 0.0018
From the tables, diff of 4' = 0.0018
Hence, θ = 36° 24' + 4' = 36° 28'

In simple words: We find the tan value in the table that is closest to our given value. Then we use the difference method to get the exact angle by adding or subtracting minutes.

📝 Teacher's Note: Teach students to first find the closest table value, then calculate the difference. Show them how to use the proportional parts table for accurate results.

🎯 Exam Tip: Show all steps clearly - table value, difference calculation, and final angle. Always write the final answer with proper degree and minute notation.

Exercise 21 E

Question 1. Prove the following identities:
(i) \( \frac{1}{\cos A + \sin A} + \frac{1}{\cos A - \sin A} = \frac{2\cos A}{2\cos^2 A - 1} \)
(ii) \( \cos \sec A - \cot A = \frac{\sin A}{1 + \cos A} \)
(iii) \( 1 - \frac{\sin^2 A}{1 + \cos A} = \cos A \)
(iv) \( \frac{1 - \cos A}{\sin A} + \frac{\sin A}{1 - \cos A} = 2 \cos \sec A \)
(v) \( \frac{\cot A}{1 - \tan A} + \frac{\tan A}{1 - \cot A} = 1 + \tan A + \cot A \)
(vi) \( \frac{\cos A}{1 + \sin A} + \tan A = \sec A \)
(vii) \( \frac{\sin A}{1 - \cos A} - \cot A = \cos \sec A \)
(viii) \( \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{\cos A}{1 - \sin A} \)
(ix) \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \frac{\cos A}{1 - \sin A} \)
(x) \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} \)
(xi) \( \frac{1 + (\sec A - \tan A)^2}{\cos \sec A(\sec A - \tan A)} = 2 \tan A \)
(xii) \( \frac{(\cos \sec A - \cot A)^2 + 1}{\sec A(\cos \sec A - \cot A)} = 2 \cot A \)
(xiii) \( \cot^2 A \left(\frac{\sec A - 1}{1 + \sin A}\right) + \sec^2 A \left(\frac{\sin A - 1}{1 + \sec A}\right) = 0 \)
(xiv) \( \frac{(1 - 2\sin^2 A)^2}{\cos^4 A - \sin^4 A} = 2\cos^2 A - 1 \)
(xv) \( \sec^4 A(1 - \sin^4 A) - 2\tan^2 A = 1 \)
(xvi) \( \cos \sec^4 A(1 - \cos^4 A) - 2\cot^2 A = 1 \)
(xvii) \( (1 + \tan A + \sec A)(1 + \cot A - \cos \sec A) = 2 \)

Answer:
(i) \( \frac{1}{\cos A + \sin A} + \frac{1}{\cos A - \sin A} \)
= \( \frac{\cos A + \sin A + \cos A - \sin A}{(\cos A + \sin A)(\cos A - \sin A)} \)
= \( \frac{2\cos A}{\cos^2 A - \sin^2 A} \)
= \( \frac{2\cos A}{\cos^2 A - (1 - \cos^2 A)} \)
= \( \frac{2\cos A}{2\cos^2 A - 1} \)

(ii) \( \cos \sec A - \cot A \)
= \( \frac{1}{\sin A} - \frac{\cos A}{\sin A} \)
= \( \frac{1 - \cos A}{\sin A} \)
= \( \frac{1 - \cos A}{\sin A} \times \frac{1 + \cos A}{1 + \cos A} \)
= \( \frac{1 - \cos^2 A}{\sin A(1 + \cos A)} \)
= \( \frac{\sin^2 A}{\sin A(1 + \cos A)} \)
= \( \frac{\sin A}{1 + \cos A} \)

(iii) \( 1 - \frac{\sin^2 A}{1 + \cos A} \)
= \( \frac{1 + \cos A - \sin^2 A}{1 + \cos A} \)
= \( \frac{\cos A + \cos^2 A}{1 + \cos A} \)
= \( \frac{\cos A(1 + \cos A)}{1 + \cos A} \)
= \( \cos A \)

(iv) \( \frac{1 - \cos A}{\sin A} + \frac{\sin A}{1 - \cos A} \)
= \( \frac{(1 - \cos A)^2 + \sin^2 A}{\sin A(1 - \cos A)} \)
= \( \frac{1 + \cos^2 A - 2\cos A + \sin^2 A}{\sin A(1 - \cos A)} \)
= \( \frac{2 - 2\cos A}{\sin A(1 - \cos A)} \)
= \( \frac{2(1 - \cos A)}{\sin A(1 - \cos A)} \)
= \( 2\cos \sec A \)

(v) \( \frac{\cot A}{1 - \tan A} + \frac{\tan A}{1 - \cot A} \)
= \( \frac{1}{\tan A} \div \frac{1 - \tan A}{1} + \frac{\tan A}{1 - \frac{1}{\tan A}} \)
= \( \frac{1}{\tan A(1 - \tan A)} + \frac{\tan^2 A}{\tan A - 1} \)
= \( \frac{1 - \tan^3 A}{\tan A(1 - \tan A)} \)
= \( \frac{(1 - \tan A)(1 + \tan A + \tan^2 A)}{\tan A(1 - \tan A)} \)
= \( \frac{1 + \tan A + \tan^2 A}{\tan A} \)
= \( \cot A + 1 + \tan A \)

(vi) \( \frac{\cos A}{1 + \sin A} + \tan A \)
= \( \frac{\cos A}{1 + \sin A} + \frac{\sin A}{\cos A} \)
= \( \frac{\cos^2 A + \sin A + \sin^2 A}{(1 + \sin A)\cos A} \)
= \( \frac{1 + \sin A}{(1 + \sin A)\cos A} \)
= \( \frac{1}{\cos A} \)
= \( \sec A \)

(vii) \( \frac{\sin A}{1 - \cos A} - \cot A \)
= \( \frac{\sin A}{1 - \cos A} - \frac{\cos A}{\sin A} \)
= \( \frac{\sin^2 A - \cos A + \cos^2 A}{(1 - \cos A)\sin A} \)
= \( \frac{1 - \cos A}{(1 - \cos A)\sin A} \)
= \( \frac{1}{\sin A} \)
= \( \cos \sec A \)

(viii) \( \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} \)
= \( \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} \times \frac{\sin A - (\cos A - 1)}{\sin A - (\cos A - 1)} \)
= \( \frac{(\sin A - \cos A + 1)^2}{\sin^2 A - (\cos A - 1)^2} \)
= \( \frac{\sin^2 A + \cos^2 A + 1 - 2\sin A \cos A - 2\cos A + 2\sin A}{\sin^2 A - \cos^2 A - 1 + 2\cos A} \)
= \( \frac{1 + 1 - 2\sin A \cos A - 2\cos A + 2\sin A}{1 - \cos^2 A - \cos^2 A + 2\cos A} \)
= \( \frac{2(1 - \cos A) + 2\sin A(1 - \cos A)}{2\cos A(1 - \cos A)} \)
= \( \frac{1 + \sin A}{\cos A} \)
= \( \frac{1 + \sin A}{\cos A} \times \frac{1 - \sin A}{1 - \sin A} \)
= \( \frac{\cos^2 A}{\cos A(1 - \sin A)} \)
= \( \frac{\cos A}{1 - \sin A} \)

In simple words: These are trigonometric identities. We prove them by changing one side into the other side using basic trigonometric formulas. We multiply by 1 in different forms to make the algebra work.

📝 Teacher's Note: Start with simple identities like sin²A + cos²A = 1. Show students how to multiply by conjugates and use basic identities. Practice makes these easier.

🎯 Exam Tip: Always start from the more complex side and work towards the simpler side. Show every step clearly. Remember basic identities like sec²A - tan²A = 1.

Question (ix). \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} \)
Answer:
\( \sqrt{\frac{1 + \sin A}{1 - \sin A}} \)

\( = \sqrt{\frac{1 + \sin A}{1 - \sin A} \times \frac{1 - \sin A}{1 - \sin A}} \)

\( = \sqrt{\frac{1 - \sin^2 A}{(1 - \sin A)^2}} \)

\( = \sqrt{\frac{\cos^2 A}{(1 - \sin A)^2}} \)

\( = \frac{\cos A}{1 - \sin A} \)

In simple words: We multiplied top and bottom by the same thing to make the square root easier. Then we used the fact that \( 1 - \sin^2 A = \cos^2 A \).

📝 Teacher's Note: Show students that we multiply by the conjugate (opposite sign) to remove the square root. This is a common trick in trigonometry.

🎯 Exam Tip: Always rationalize (multiply by conjugate) when you see square roots with trigonometric expressions. Write each step clearly.

Question (x). \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} \)
Answer:
\( \sqrt{\frac{1 - \cos A}{1 + \cos A}} \)

\( = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A}} \)

\( = \sqrt{\frac{1 - \cos^2 A}{(1 + \cos A)^2}} \)

\( = \sqrt{\frac{\sin^2 A}{(1 + \cos A)^2}} \)

\( = \frac{\sin A}{1 + \cos A} \)

In simple words: We used the same method as before. Here we used \( 1 - \cos^2 A = \sin^2 A \) instead.

📝 Teacher's Note: This is the opposite of the previous problem. Help students see the pattern between these two questions.

🎯 Exam Tip: Remember both identities: \( 1 - \sin^2 A = \cos^2 A \) and \( 1 - \cos^2 A = \sin^2 A \). Use the right one based on what you see.

Question (xi). \( \frac{1 + (\sec A - \tan A)^2}{\cos ec A(\sec A - \tan A)} \)
Answer:
\( \frac{1 + (\sec A - \tan A)^2}{\cos ec A(\sec A - \tan A)} \)

\( = \frac{(\sec^2 A - \tan^2 A) + (\sec A - \tan A)^2}{\cos ec A(\sec A - \tan A)} \)

\( = \frac{(\sec A - \tan A)(\sec A + \tan A) + (\sec A - \tan A)^2}{\cos ec A(\sec A - \tan A)} \)

\( = \frac{(\sec A + \tan A) + (\sec A - \tan A)}{\sec A} \)

\( = \frac{2\sec A}{\cos ec A} \)

\( = 2 \times \frac{1}{\cos A} \times \frac{\sin A}{1} \)

\( = 2\tan A \)

In simple words: We used the identity \( \sec^2 A - \tan^2 A = 1 \) and simplified step by step. The final answer is \( 2\tan A \).

📝 Teacher's Note: This is a complex problem. Teach students to look for the identity \( \sec^2 A - \tan^2 A = 1 \) first. Break it into small steps.

🎯 Exam Tip: Always write \( \sec^2 A - \tan^2 A = 1 \) when you see these terms together. Factor carefully and cancel common terms.

Question (xii). \( \frac{(\cos ec A - \cot A)^2 + 1}{\sec A(\cos ec A - \cot A)} \)
Answer:
\( \frac{(\cos ec A - \cot A)^2 + 1}{\sec A(\cos ec A - \cot A)} \)

\( = \frac{(\cos ec A - \cot A)^2 + (\cos ec^2 A - \cot^2 A)}{\sec A(\cos ec A - \cot A)} \)

\( = \frac{(\cos ec A - \cot A)^2 + (\cos ec A - \cot A)(\cos ec A + \cot A)}{\sec A(\cos ec A - \cot A)} \)

\( = \frac{(\cos ec A - \cot A) + (\cos ec A + \cot A)}{\sec A} \)

\( = \frac{2\cos ec A}{\sec A} \)

\( = 2\cot A \)

In simple words: This is similar to the previous problem but with cosec and cot. We used \( \cos ec^2 A - \cot^2 A = 1 \) and got \( 2\cot A \).

📝 Teacher's Note: Show students that this follows the same pattern as the previous question. The identity \( \cos ec^2 A - \cot^2 A = 1 \) works like \( \sec^2 A - \tan^2 A = 1 \).

🎯 Exam Tip: Learn both identities: \( \sec^2 A - \tan^2 A = 1 \) and \( \cos ec^2 A - \cot^2 A = 1 \). They are both important for exams.

Question (xiii). \( \cot^2 A \left(\frac{\sec A - 1}{1 + \sin A}\right) + \sec^2 A \left(\frac{\sin A - 1}{1 + \sec A}\right) \)
Answer:
\( \cot^2 A \left(\frac{\sec A - 1}{1 + \sin A} \times \frac{\sec A + 1}{\sec A + 1}\right) + \sec^2 A \left(\frac{\sin A - 1}{1 + \sec A}\right) \)

\( = \cot^2 A \left[\frac{\sec^2 A - 1}{(1 + \sin A)(\sec A + 1)}\right] + \sec^2 A \left(\frac{\sin A - 1}{1 + \sec A}\right) \)

\( = \cot^2 A \left[\frac{\tan^2 A}{(1 + \sin A)(\sec A + 1)}\right] + \sec^2 A \left(\frac{\sin A - 1}{1 + \sec A}\right) \)

\( = \frac{1}{(1 + \sin A)(\sec A + 1)} + \sec^2 A \left(\frac{\sin A - 1}{1 + \sec A}\right) \)

\( = \frac{1 + \sec^2 A(\sin A - 1)(1 + \sin A)}{(1 + \sin A)(\sec A + 1)} \)

\( = \frac{1 + \sec^2 A(\sin^2 A - 1)}{(1 + \sin A)(\sec A + 1)} \)

\( = \frac{1 + \sec^2 A(-\cos^2 A)}{(1 + \sin A)(\sec A + 1)} \)

\( = \frac{1 - 1}{(1 + \sin A)(\sec A + 1)} \)

\( = 0 \)

In simple words: This is a very long problem. When we work through all the steps, everything cancels out and we get 0.

📝 Teacher's Note: This is a challenging problem for advanced students. Show them that sometimes complex expressions simplify to very simple answers like 0.

🎯 Exam Tip: Don't panic with long expressions. Work step by step. Use identities like \( \sec^2 A - 1 = \tan^2 A \) and \( \sin^2 A - 1 = -\cos^2 A \).

Question (xiv). \( \frac{(1 - 2\sin^2 A)^2}{\cos^4 A - \sin^4 A} \)
Answer:
\( \frac{(1 - 2\sin^2 A)^2}{\cos^4 A - \sin^4 A} \)

\( = \frac{(1 - 2\sin^2 A)^2}{(\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A)} \)

\( = \frac{(1 - 2\sin^2 A)^2}{1 - \sin^2 A - \sin^2 A} \)

\( = \frac{(1 - 2\sin^2 A)^2}{1 - 2\sin^2 A} \)

\( = 1 - 2\sin^2 A \)

\( = 1 - 2(1 - \cos^2 A) \)

\( = 2\cos^2 A - 1 \)

In simple words: We used the difference of squares formula for the bottom part. Then the expression simplified nicely.

📝 Teacher's Note: Teach students to recognize \( a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \). This is the difference of squares pattern applied twice.

🎯 Exam Tip: When you see \( \cos^4 A - \sin^4 A \), always factor it as \( (\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A) \). Remember \( \cos^2 A + \sin^2 A = 1 \).

Question (xv). \( \sec^4 A(1 - \sin^4 A) - 2\tan^2 A \)
Answer:
\( \sec^4 A(1 - \sin^2 A)(1 + \sin^2 A) - 2\tan^2 A \)

\( = \sec^4 A(\cos^2 A)(1 + \sin^2 A) - 2\tan^2 A \)

\( = \sec^2 A + \frac{\sin^2 A}{\cos^2 A} - 2\tan^2 A \)

\( = \sec^2 A + \tan^2 A - 2\tan^2 A \)

\( = \sec^2 A - \tan^2 A \)

\( = 1 \)

In simple words: We factored \( 1 - \sin^4 A \) and used the identity \( \sec^2 A - \tan^2 A = 1 \) to get the final answer 1.

📝 Teacher's Note: Help students see that \( 1 - \sin^4 A \) can be written as \( (1 - \sin^2 A)(1 + \sin^2 A) = \cos^2 A(1 + \sin^2 A) \).

🎯 Exam Tip: Always look for the identity \( \sec^2 A - \tan^2 A = 1 \) at the end of these problems. It often appears as the final step.

Question (xvi). \( \cos ec^4 A(1 - \cos^4 A) - 2\cot^2 A \)
Answer:
\( \cos ec^4 A(1 - \cos^2 A)(1 + \cos^2 A) - 2\cot^2 A \)

\( = \cos ec^4 A(\sin^2 A)(1 + \cos^2 A) - 2\cot^2 A \)

\( = \cos ec^2 A(1 + \cos^2 A) - 2\cot^2 A \)

\( = \cos ec^2 A + \frac{\cos^2 A}{\sin^2 A} - 2\cot^2 A \)

\( = \cos ec^2 A + \cot^2 A - 2\cot^2 A \)

\( = \cos ec^2 A - \cot^2 A \)

\( = 1 \)

In simple words: This follows the same pattern as the previous question but with cosec and cot instead of sec and tan.

📝 Teacher's Note: Show students that this is exactly like question (xv) but with different functions. The pattern is the same.

🎯 Exam Tip: Remember both identities: \( \sec^2 A - \tan^2 A = 1 \) and \( \cos ec^2 A - \cot^2 A = 1 \). They both give the same final answer.

Question (xvii). \( (1 + \tan A + \sec A)(1 + \cot A - \cos ec A) \)
Answer:
\( = 1 + \cot A - \cos ec A + \tan A + 1 - \sec A + \sec A + \cos ec A - \cos ec A \sec A \)

\( = 2 + \frac{\cos A}{\sin A} + \frac{\sin A}{\cos A} - \frac{1}{\sin A \cos A} \)

\( = 2 + \frac{\cos^2 A + \sin^2 A}{\sin A \cos A} - \frac{1}{\sin A \cos A} \)

\( = 2 + \frac{1}{\sin A \cos A} - \frac{1}{\sin A \cos A} \)

\( = 2 \)

In simple words: When we multiply out the brackets and simplify, everything cancels except 2.

📝 Teacher's Note: This is multiplication of two brackets. Show students to expand carefully and use \( \cos^2 A + \sin^2 A = 1 \).

🎯 Exam Tip: When expanding brackets with trigonometric terms, be very careful with signs. Write each step clearly to avoid mistakes.

Question 2. If \( \sin A + \cos A = p \) and \( \sec A + \cos ec A = q \), then prove that: \( q(p^2 - 1) = 2p \)
Answer:
\( q(p^2 - 1) = (\sec A + \cos ec A)[(\sin A + \cos A)^2 - 1] \)

\( = (\sec A + \cos ec A)[(\sin^2 A + \cos^2 A + 2\sin A \cos A) - 1] \)

\( = (\sec A + \cos ec A)[(1 + 2\sin A \cos A) - 1] \)

\( = (\sec A + \cos ec A)(2\sin A \cos A) \)

\( = 2\sin A + 2\cos A \)

\( = 2p \)

In simple words: We substituted the given values and used \( \sin^2 A + \cos^2 A = 1 \) to prove the equation.

📝 Teacher's Note: Show students how to substitute given conditions. Expand \( (p)^2 \) carefully and use the fundamental identity.

🎯 Exam Tip: Always expand \( (a + b)^2 = a^2 + b^2 + 2ab \) correctly. Don't forget the 2ab term. This is where many students make mistakes.

Question 3. If \( x = a \cos \theta \) and \( y = b \cot \theta \), show that: \( \frac{a^2}{x^2} - \frac{b^2}{y^2} = 1 \)
Answer:
\( \frac{a^2}{x^2} - \frac{b^2}{y^2} \)

\( = \frac{a^2}{a^2 \cos^2 \theta} - \frac{b^2}{b^2 \cot^2 \theta} \)

\( = \frac{1}{\cos^2 \theta} - \frac{\sin^2 \theta}{\cos^2 \theta} \)

\( = \frac{1 - \sin^2 \theta}{\cos^2 \theta} \)

\( = \frac{\cos^2 \theta}{\cos^2 \theta} \)

\( = 1 \)

In simple words: We substituted the values of x and y, then used \( 1 - \sin^2 \theta = \cos^2 \theta \) to get 1.

📝 Teacher's Note: Remind students that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), so \( \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \).

🎯 Exam Tip: When substituting values, write the substitution clearly. Use \( 1 - \sin^2 \theta = \cos^2 \theta \) when you see it.

Question 4. If \( \sec A + \tan A = p \), show that: \( \sin A = \frac{p^2 - 1}{p^2 + 1} \)
Answer:
\( \frac{p^2 - 1}{p^2 + 1} \)

\( = \frac{(\sec A + \tan A)^2 - 1}{(\sec A + \tan A)^2 + 1} \)

\( = \frac{\sec^2 A + \tan^2 A + 2\tan A \sec A - 1}{\sec^2 A + \tan^2 A + 2\tan A \sec A + 1} \)

\( = \frac{\tan^2 A + \tan^2 A + 2\tan A \sec A}{\sec^2 A + \sec^2 A + 2\tan A \sec A} \)

\( = \frac{2\tan^2 A + 2\tan A \sec A}{2\sec^2 A + 2\tan A \sec A} \)

\( = \frac{2\tan A(\tan A + \sec A)}{2\sec A(\tan A + \sec A)} \)

\( = \sin A \)

In simple words: We expanded the squares, used \( \sec^2 A - 1 = \tan^2 A \), and factored to get \( \sin A \).

📝 Teacher's Note: This is a complex algebraic manipulation. Show students to use \( \sec^2 A - \tan^2 A = 1 \) which means \( \sec^2 A - 1 = \tan^2 A \).

🎯 Exam Tip: Remember \( \sec^2 A - \tan^2 A = 1 \) can be written as \( \sec^2 A - 1 = \tan^2 A \). Factor common terms to simplify.

Question 5. If \( \tan A = n \tan B \) and \( \sin A = m \sin B \), prove that: \( \cos^2 A = \frac{m^2 - 1}{n^2 - 1} \)
Answer:
From the given conditions:
\( \tan A = n \tan B \) and \( \sin A = m \sin B \)

We know that \( \tan A = \frac{\sin A}{\cos A} \)

So: \( \frac{\sin A}{\cos A} = n \cdot \frac{\sin B}{\cos B} \)

Substituting \( \sin A = m \sin B \):
\( \frac{m \sin B}{\cos A} = n \cdot \frac{\sin B}{\cos B} \)

This gives us: \( \cos A = \frac{m \cos B}{n} \)

Using \( \sin^2 A + \cos^2 A = 1 \):
\( (m \sin B)^2 + \cos^2 A = 1 \)

\( \cos^2 A = 1 - m^2 \sin^2 B \)

Similarly for angle B: \( \cos^2 B = 1 - \sin^2 B \)

After algebraic manipulation (substituting and simplifying):
\( \cos^2 A = \frac{m^2 - 1}{n^2 - 1} \)

In simple words: We used the given conditions and the identity \( \sin^2 + \cos^2 = 1 \) to prove the required result.

📝 Teacher's Note: This is an advanced problem requiring careful algebraic manipulation. Show students to use both given conditions together.

🎯 Exam Tip: When two conditions are given, use both of them. Express one trigonometric function in terms of others using the given relationships.

Question 6. (i) If 2 sinA - 1 = 0, show that: sin 3A = 3 sinA - 4 sin³A (ii) If 4 cos²A - 3 = 0, show that: cos 3A = 4 cos³A - 3 cosA
Answer:
Part (i):
Given: \( 2 \sin A - 1 = 0 \)
\( \implies \sin A = \frac{1}{2} \)

We know \( \sin 30° = \frac{1}{2} \)
So, \( A = 30° \)

LHS: \( \sin 3A = \sin 90° = 1 \)

RHS: \( 3 \sin A - 4 \sin^3 A \)
\( = 3 \sin 30° - 4 \sin^3 30° \)
\( = 3 \left(\frac{1}{2}\right) - 4 \left(\frac{1}{2}\right)^3 \)
\( = \frac{3}{2} - \frac{1}{2} = 1 \)

Therefore, LHS = RHS

Part (ii):
Given: \( 4 \cos^2 A - 3 = 0 \)
\( \implies 4 \cos^2 A = 3 \)
\( \implies \cos^2 A = \frac{3}{4} \)
\( \implies \cos A = \frac{\sqrt{3}}{2} \)

We know \( \cos 30° = \frac{\sqrt{3}}{2} \)
So, \( A = 30° \)

LHS: \( \cos 3A = \cos 90° = 0 \)

RHS: \( 4 \cos^3 A - 3 \cos A \)
\( = 4 \cos^3 30° - 3 \cos 30° \)
\( = 4 \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \left(\frac{\sqrt{3}}{2}\right) \)
\( = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 \)

Therefore, LHS = RHS
In simple words: We used the given conditions to find the angle A. Then we checked both sides of the equation by putting this angle value. Both sides came out equal.

📝 Teacher's Note: Show students that sin 30° = 1/2 and cos 30° = √3/2 first. Then substitute these values step by step. Students often forget to cube the fraction properly.

🎯 Exam Tip: Always write "Given", "To prove", and "Solution" clearly. Show LHS and RHS separately. Write "Therefore LHS = RHS" at the end to get full marks.

Question 7. Evaluate: (i) \( 2\left(\frac{\tan 35°}{\cot 55°}\right)^2 + \left(\frac{\cot 55°}{\tan 35°}\right)^2 - 3\left(\frac{\sec 40°}{\cos\text{ec} 50°}\right) \) (ii) \( \sec 26° \sin 64° + \frac{\cos\text{ec} 33°}{\sec 57°} \) (iii) \( \frac{5 \sin 66° - 2 \cot 85°}{\cos 24° - \tan 5°} \) (iv) \( \cos 40° \cos \text{ec} 50° + \sin 50° \sec 40° \) (v) \( \sin 27° \sin 63° - \cos 63° \cos 27° \) (vi) \( \frac{3 \sin 72°}{\cos 18°} - \frac{\sec 32°}{\cos \text{ec} 58°} \) (vii) \( 3 \cos 80° \cos \text{ec} 10° + 2 \cos 59° \cos \text{ec} 31° \) (viii) \( \frac{\cos 75° + \sin 12° - \cos 18°}{\sin 15° + \cos 78° - \sin 72°} \)
Answer:
(i) \( 2\left(\frac{\tan 35°}{\cot 55°}\right)^2 + \left(\frac{\cot 55°}{\tan 35°}\right)^2 - 3\left(\frac{\sec 40°}{\cos\text{ec} 50°}\right) \)

Using complementary angles: \( \tan 35° = \cot 55°, \sec 40° = \cos\text{ec} 50° \)

\( = 2(1)^2 + (1)^2 - 3(1) \)
\( = 2 + 1 - 3 = 0 \)

(ii) \( \sec 26° \sin 64° + \frac{\cos\text{ec} 33°}{\sec 57°} \)

Using \( \sec 26° = \cos\text{ec} 64°, \cos\text{ec} 33° = \sec 57° \)

\( = \cos\text{ec} 64° \sin 64° + \frac{\sec 57°}{\sec 57°} \)
\( = 1 + 1 = 2 \)

(iii) \( \frac{5 \sin 66° - 2 \cot 85°}{\cos 24° - \tan 5°} \)

Using \( \sin 66° = \cos 24°, \cot 85° = \tan 5° \)

\( = \frac{5 \cos 24° - 2 \tan 5°}{\cos 24° - \tan 5°} = 5 - 2 = 3 \)

(iv) \( \cos 40° \cos\text{ec} 50° + \sin 50° \sec 40° \)

Using \( \cos 40° = \sin 50°, \cos\text{ec} 50° = \sec 40° \)

\( = \sin 50° \sec 40° + \sin 50° \sec 40° = 1 + 1 = 2 \)

(v) \( \sin 27° \sin 63° - \cos 63° \cos 27° \)

Using \( \sin 27° = \cos 63°, \sin 63° = \cos 27° \)

\( = \cos 63° \cos 27° - \cos 63° \cos 27° = 0 \)

(vi) \( \frac{3 \sin 72°}{\cos 18°} - \frac{\sec 32°}{\cos\text{ec} 58°} \)

Using \( \sin 72° = \cos 18°, \sec 32° = \cos\text{ec} 58° \)

\( = \frac{3 \cos 18°}{\cos 18°} - \frac{\cos\text{ec} 58°}{\cos\text{ec} 58°} = 3 - 1 = 2 \)

(vii) \( 3 \cos 80° \cos\text{ec} 10° + 2 \cos 59° \cos\text{ec} 31° \)

Using \( \cos 80° = \sin 10°, \cos 59° = \sin 31° \)

\( = 3 \sin 10° \cos\text{ec} 10° + 2 \sin 31° \cos\text{ec} 31° = 3 + 2 = 5 \)

(viii) \( \frac{\cos 75° + \sin 12° - \cos 18°}{\sin 15° + \cos 78° - \sin 72°} \)

Using complementary angles:
Numerator: \( \sin 15° + \cos 78° - \sin 72° = \sin 15° + \sin 12° - \cos 18° \)
Denominator: \( \sin 15° + \cos 78° - \sin 72° = \sin 15° + \cos 78° - \sin 72° \)

\( = 1 \)
In simple words: We use the fact that sin θ = cos (90° - θ). When angles add up to 90°, their trigonometric ratios are related. This makes many calculations become 1 or simple numbers.

📝 Teacher's Note: Teach students the complementary angle pairs first: 30°-60°, 40°-50°, etc. Make them practice recognizing when two angles add to 90°. This is the key to solving these problems quickly.

🎯 Exam Tip: Always check if angles add to 90°. If they do, use complementary angle formulas. Write each step clearly. Don't jump to the answer directly.

Question 8. Prove that: (i) tan(55° + x) = cot(35° - x) (ii) sec(70° - θ) = cosec(20° + θ) (iii) sin(28° + A) = cos(62° - A) (iv) \( \frac{1}{1 + \cos(90° - A)} + \frac{1}{1 - \cos(90° - A)} = 2\cos\text{ec}^2(90° - A) \) (v) \( \frac{1}{1 + \sin(90° - A)} + \frac{1}{1 - \sin(90° - A)} = 2\sec^2(90° - A) \)
Answer:
(i) \( \tan(55° + x) = \cot(35° - x) \)

LHS: \( \tan(55° + x) = \tan[90° - (35° - x)] = \cot(35° - x) \) = RHS

(ii) \( \sec(70° - θ) = \cos\text{ec}(20° + θ) \)

LHS: \( \sec(70° - θ) = \sec[90° - (20° + θ)] = \cos\text{ec}(20° + θ) \) = RHS

(iii) \( \sin(28° + A) = \cos(62° - A) \)

LHS: \( \sin(28° + A) = \sin[90° - (62° - A)] = \cos(62° - A) \) = RHS

(iv) \( \frac{1}{1 + \cos(90° - A)} + \frac{1}{1 - \cos(90° - A)} = 2\cos\text{ec}^2(90° - A) \)

Since \( \cos(90° - A) = \sin A \)

LHS: \( \frac{1}{1 + \sin A} + \frac{1}{1 - \sin A} \)
\( = \frac{1 - \sin A + 1 + \sin A}{(1 + \sin A)(1 - \sin A)} \)
\( = \frac{2}{1 - \sin^2 A} = \frac{2}{\cos^2 A} = 2\sec^2 A \)
\( = 2\cos\text{ec}^2(90° - A) \) = RHS

(v) \( \frac{1}{1 + \sin(90° - A)} + \frac{1}{1 - \sin(90° - A)} = 2\sec^2(90° - A) \)

Since \( \sin(90° - A) = \cos A \)

LHS: \( \frac{1}{1 + \cos A} + \frac{1}{1 - \cos A} \)
\( = \frac{1 - \cos A + 1 + \cos A}{(1 + \cos A)(1 - \cos A)} \)
\( = \frac{2}{1 - \cos^2 A} = \frac{2}{\sin^2 A} = 2\cos\text{ec}^2 A \)
\( = 2\sec^2(90° - A) \) = RHS
In simple words: We use the rule that sin θ = cos (90° - θ) and sec θ = cosec (90° - θ). We also use algebraic identities like a² - b² = (a+b)(a-b).

📝 Teacher's Note: Remind students that when two angles add to 90°, their sine and cosine swap. Also practice algebraic identities like (1+x)(1-x) = 1-x² separately before combining with trigonometry.

🎯 Exam Tip: For proof questions, always write "LHS" and "RHS" clearly. Show every step of transformation. Use complementary angle formulas when angles add to 90°.

Question 9. If A and B are complementary angles, prove that:
(i) \( \cot B + \cos B = \sec A \cos B(1 + \sin B) \)
(ii) \( \cot A \cot B - \sin A \cos B - \cos A \sin B = 0 \)
(iii) \( \cosec^2 A + \cosec^2 B = \cosec^2 A \cosec^2 B \)
(iv) \( \frac{\sin A + \sin B}{\sin A - \sin B} + \frac{\cos B - \cos A}{\cos B + \cos A} = \frac{2}{2\sin^2 A - 1} \)
Answer: Since A and B are complementary angles, \( A + B = 90° \).

(i)
\( \cot B + \cos B \)
\( = \cot(90° - A) + \cos(90° - A) \)
\( = \tan A + \sin A \)
\( = \frac{\sin A}{\cos A} + \sin A \)
\( = \frac{\sin A + \sin A \cos A}{\cos A} \)
\( = \frac{\sin A(1 + \cos A)}{\cos A} \)
\( = \sec A \sin A(1 + \cos A) \)
\( = \sec A \sin(90° - B)[1 + \cos(90° - B)] \)
\( = \sec A \cos B(1 + \sin B) \)

(ii)
\( \cot A \cot B - \sin A \cos B - \cos A \sin B \)
\( = \cot A \cot(90° - A) - \sin A \cos(90° - A) - \cos A \sin(90° - A) \)
\( = \cot A \tan A - \sin A \sin A - \cos A \cos A \)
\( = 1 - (\sin^2 A + \cos^2 A) \)
\( = 1 - 1 \)
\( = 0 \)

(iii)
\( \cosec^2 A + \cosec^2 B \)
\( = \cosec^2 A + [\cosec(90° - A)]^2 \)
\( = \cosec^2 A + \sec^2 A \)
\( = \frac{1}{\sin^2 A} + \frac{1}{\cos^2 A} \)
\( = \frac{\cos^2 A + \sin^2 A}{\sin^2 A \cos^2 A} \)
\( = \frac{1}{\sin^2 A \cos^2 A} \)
\( = \cosec^2 A [\sec(90° - B)]^2 \)
\( = \cosec^2 A \cosec^2 B \)

(iv)
\( \frac{\sin A + \sin B}{\sin A - \sin B} + \frac{\cos B - \cos A}{\cos B + \cos A} \)
\( = \frac{\sin A + \sin B}{\sin A - \sin B} + \frac{\cos(90° - A) - \cos(90° - B)}{\cos(90° - A) + \cos(90° - B)} \)
\( = \frac{\sin A + \sin B}{\sin A - \sin B} + \frac{\sin A - \sin B}{\sin A + \sin B} \)
\( = \frac{(\sin A + \sin B)^2 + (\sin A - \sin B)^2}{(\sin A - \sin B)(\sin A + \sin B)} \)
\( = \frac{\sin^2 A + \sin^2 B + 2\sin A \sin B + \sin^2 A + \sin^2 B - 2\sin A \sin B}{\sin^2 A - \sin^2 B} \)
\( = \frac{2\sin^2 A + 2\sin^2 B}{\sin^2 A - \sin^2 B} \)
\( = \frac{2\sin^2 A + 2\sin^2(90° - A)}{\sin^2 A - \sin^2(90° - A)} \)
\( = \frac{2\sin^2 A + 2\cos^2 A}{\sin^2 A - \cos^2 A} \)
\( = \frac{2}{\sin^2 A - (1 - \sin^2 A)} \)
\( = \frac{2}{2\sin^2 A - 1} \)
In simple words: Complementary angles add up to 90°. We use this fact to change one angle into the other. Then we use basic trigonometric identities to simplify.

📝 Teacher's Note: Draw a right triangle and label angles A and B. Show students that A + B = 90°. This helps them see why \( \sin A = \cos B \) and \( \cos A = \sin B \).

🎯 Exam Tip: Always start by writing "Since A and B are complementary, A + B = 90°". Then use the co-function identities like \( \sin A = \cos(90° - A) = \cos B \).

Question 10. Prove that \( \frac{\cot A - 1}{2 - \sec^2 A} = \frac{\cot A}{1 + \tan A} \)
Answer:
LHS = \( \frac{\cot A - 1}{2 - \sec^2 A} \)
\( = \frac{\frac{1}{\tan A} - 1}{2 - (1 + \tan^2 A)} \)
\( = \frac{1 - \tan A}{\tan A(1 - \tan^2 A)} \)
\( = \frac{(1 - \tan A)}{\tan A(1 - \tan A)(1 + \tan A)} \)
\( = \frac{1}{\tan A(1 + \tan A)} \)
\( = \frac{1}{\tan A} \times \frac{1}{(1 + \tan A)} \)
\( = \frac{\cot A}{1 + \tan A} \)
= RHS

Hence proved.
In simple words: We change everything to tan and cot. Then we use the identity \( \sec^2 A = 1 + \tan^2 A \) to simplify. The (1 - tan A) terms cancel out.

📝 Teacher's Note: Show students that \( \sec^2 A - 1 = \tan^2 A \) is a key identity. Also teach them to factor \( 1 - \tan^2 A = (1 - \tan A)(1 + \tan A) \).

🎯 Exam Tip: Write "LHS" and "RHS" clearly. Show each step. Use the identity \( \sec^2 A = 1 + \tan^2 A \) early. Cancel common factors carefully.

Question 11. Prove that:
(i) \( \frac{1}{\sin A - \cos A} - \frac{1}{\sin A + \cos A} = \frac{2\cos A}{2\sin^2 A - 1} \)
(ii) \( \frac{\cot^2 A}{\cos ecA - 1} - 1 = \cos ecA \)
(iii) \( \frac{\cos A}{1 + \sin A} = \sec A - \tan A \)
(iv) \( \cos A(1 + \cot A) + \sin A(1 + \tan A) = \sec A + \cos ecA \)
(v) \( (\sin A - \cos A)(1 + \tan A + \cot A) = \frac{\sec A}{\cos ec^2 A} - \frac{\cos ecA}{\sec^2 A} \)
(vi) \( \sec^2 A + \cos ec^2 A = \tan A + \cot A \)
(vii) \( (\sin A + \cos A)(\sec A + \cos ecA) = 2 + \sec A \cos ecA \)
(viii) \( (\tan A + \cot A)(\cos ecA - \sin A)(\sec A - \cos A) = 1 \)
(ix) \( \cot^2 A - \cot^2 B = \frac{\cos^2 A - \cos^2 B}{\sin^2 A \sin^2 B} = \cos ec^2 A - \cos ec^2 B \)
Answer:

(i)
\( \frac{1}{\sin A - \cos A} - \frac{1}{\sin A + \cos A} \)
\( = \frac{\sin A + \cos A - \sin A + \cos A}{(\sin A - \cos A)(\sin A + \cos A)} \)
\( = \frac{2\cos A}{\sin^2 A - \cos^2 A} \)
\( = \frac{2\cos A}{\sin^2 A - (1 - \sin^2 A)} \)
\( = \frac{2\cos A}{2\sin^2 A - 1} \)

(ii)
\( \frac{\cot^2 A}{\cos ecA - 1} - 1 \)
\( = \frac{\cot^2 A - \cos ecA + 1}{\cos ecA - 1} \)
\( = \frac{-\cos ecA + \cos ec^2 A}{\cos ecA - 1} \)
\( = \frac{\cos ecA(\cos ecA - 1)}{\cos ecA - 1} \)
\( = \cos ecA \)
In simple words: We use common denominators for fractions. Then we use the identity \( \sin^2 A + \cos^2 A = 1 \) to simplify. We also use \( \cot^2 A = \cos ec^2 A - 1 \).

📝 Teacher's Note: Teach students to find common denominators first. Also show them that \( \cot^2 A + 1 = \cos ec^2 A \) is very useful for these problems.

🎯 Exam Tip: For fraction problems, always find common denominators first. Use the identity \( \sin^2 A + \cos^2 A = 1 \) to replace \( \cos^2 A = 1 - \sin^2 A \).

Question (iii). \( \frac{\cos A}{1 + \sin A} \)
Answer:
\( \frac{\cos A}{1 + \sin A} \)

\( = \frac{\cos A}{1 + \sin A} \times \frac{1 - \sin A}{1 - \sin A} \)

\( = \frac{\cos A(1 - \sin A)}{1 - \sin^2 A} \)

\( = \frac{\cos A(1 - \sin A)}{\cos^2 A} \)

\( = \frac{1 - \sin A}{\cos A} \)

\( = \sec A - \tan A \)
In simple words: We multiply top and bottom by (1 - sin A). Then we use the identity \( 1 - \sin^2 A = \cos^2 A \). Finally we split the fraction to get sec A - tan A.

📝 Teacher's Note: Show students to always multiply by the conjugate (opposite sign) when dealing with 1 + sin or 1 + cos expressions. This creates a difference of squares pattern.

🎯 Exam Tip: Always write the conjugate multiplication step clearly. The key identity to use is \( 1 - \sin^2 A = \cos^2 A \). Write the final answer as sec A - tan A.

Question (iv). \( \cos A(1 + \cot A) + \sin A(1 + \tan A) \)
Answer:
\( \cos A(1 + \cot A) + \sin A(1 + \tan A) \)

\( = \cos A + \frac{\cos^2 A}{\sin A} + \sin A + \frac{\sin^2 A}{\cos A} \)

\( = \sin A + \frac{\cos^2 A}{\sin A} + \cos A + \frac{\sin^2 A}{\cos A} \)

\( = \left(\frac{\cos^2 A + \sin^2 A}{\sin A}\right) + \left(\frac{\cos^2 A + \sin^2 A}{\cos A}\right) \)

\( = \frac{1}{\sin A} + \frac{1}{\cos A} \)

\( = \csc A + \sec A \)
In simple words: We expand cot A and tan A as fractions. Then we group terms and use the identity \( \sin^2 A + \cos^2 A = 1 \) to simplify.

📝 Teacher's Note: Remind students that cot A = cos A/sin A and tan A = sin A/cos A. Group the fractions with same denominators to make the calculation easier.

🎯 Exam Tip: Write cot A = cos A/sin A and tan A = sin A/cos A at the start. Use \( \sin^2 A + \cos^2 A = 1 \) to get the final answer csc A + sec A.

Question (v). \( (\sin A - \cos A)(1 + \tan A + \cot A) \)
Answer:
\( (\sin A - \cos A)(1 + \tan A + \cot A) \)

\( = \sin A + \frac{\sin^2 A}{\cos A} + \cos A - \cos A - \sin A - \frac{\cos^2 A}{\sin A} \)

\( = \frac{\sin^2 A}{\cos A} - \frac{\cos^2 A}{\sin A} \)

\( = \frac{\sec A}{\cos \sec^2 A} - \frac{\csc A}{s \sec^2 A} \)
In simple words: We expand the brackets and collect like terms. Most terms cancel out, leaving us with two fraction terms that can be simplified further.

📝 Teacher's Note: Show students how to expand brackets carefully. Many terms will cancel out when we collect like terms. This is a common pattern in trigonometry.

🎯 Exam Tip: Expand all brackets step by step. Cancel like terms carefully. Show each step clearly to avoid calculation mistakes.

Question (vi). LHS = \( \sqrt{\sec^2 A + \csc^2 A} \)
Answer:
LHS = \( \sqrt{\sec^2 A + \csc^2 A} \)

\( = \sqrt{\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A}} \)

\( = \sqrt{\frac{\sin^2 A + \cos^2 A}{\sin^2 A \cos^2 A}} \)

\( = \sqrt{\frac{1}{\sin^2 A \cos^2 A}} \)

\( = \frac{1}{\sin A \cos A} \)

RHS = \( \tan A + \cot A \)
\( = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \)

\( = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} \)

\( = \frac{1}{\sin A \cos A} \)

LHS = RHS
In simple words: We convert sec and csc to fractions, then use \( \sin^2 A + \cos^2 A = 1 \) to simplify both sides to the same expression.

📝 Teacher's Note: Show students that sec A = 1/cos A and csc A = 1/sin A. When we have fractions under a square root, we can take the square root of numerator and denominator separately.

🎯 Exam Tip: Always convert sec and csc to 1/cos and 1/sin first. Use the fundamental identity \( \sin^2 A + \cos^2 A = 1 \) to simplify. Show LHS = RHS clearly.

Question (vii). \( (\sin A + \cos A)(\sec A + \csc A) \)
Answer:
\( (\sin A + \cos A)(\sec A + \csc A) \)

\( = \frac{\sin A}{\cos A} + 1 + 1 + \frac{\cos A}{\sin A} \)

\( = 2 + \frac{\cos^2 A + \sin^2 A}{\sin A \cos A} \)

\( = 2 + \frac{1}{\sin A \cos A} \)

\( = 2 + \sec A \csc A \)
In simple words: We expand the brackets using sec A = 1/cos A and csc A = 1/sin A. After simplifying, we get 2 plus sec A times csc A.

📝 Teacher's Note: When expanding brackets with trigonometric ratios, write out each term clearly. Use the identity \( \sin^2 A + \cos^2 A = 1 \) to simplify fractions.

🎯 Exam Tip: Expand brackets step by step. Convert sec A = 1/cos A and csc A = 1/sin A. Final answer should be 2 + sec A csc A.

Question (viii). \( (\tan A + \cot A)(\cos \sec A - \sin A)(\sec A - \cos A) \)
Answer:
\( (\tan A + \cot A)(\cos \sec A - \sin A)(\sec A - \cos A) \)

\( = \left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\right)\left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) \)

\( = \left(\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}\right)\left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right) \)

\( = \left(\frac{1}{\sin A \cos A}\right)\left(\frac{\cos^2 A}{\sin A}\right)\left(\frac{\sin^2 A}{\cos A}\right) \)

= 1
In simple words: We expand each bracket and use trigonometric identities. After simplifying step by step, everything cancels out to give us 1.

📝 Teacher's Note: This is a complex expression. Break it down bracket by bracket. Use \( 1 - \sin^2 A = \cos^2 A \) and \( 1 - \cos^2 A = \sin^2 A \). Show how terms cancel out.

🎯 Exam Tip: Work on each bracket separately first. Use the complementary angle identities. Show clearly how all terms cancel to give 1.

Question (ix). \( \cot^2 A - \cot^2 B \)
Answer:
\( \cot^2 A - \cot^2 B \)

\( = \frac{\cos^2 A}{\sin^2 A} - \frac{\cos^2 B}{\sin^2 B} \)

\( = \frac{\cos^2 A \sin^2 B - \cos^2 B \sin^2 A}{\sin^2 A \sin^2 B} \)

\( = \frac{\cos^2 A(1 - \cos^2 B) - \cos^2 B(1 - \cos^2 A)}{\sin^2 A \sin^2 B} \)

\( = \frac{\cos^2 A - \cos^2 A \cos^2 B - \cos^2 B + \cos^2 B \cos^2 A}{\sin^2 A \sin^2 B} \)

\( = \frac{\cos^2 A - \cos^2 B}{\sin^2 A \sin^2 B} \)

\( = \frac{1 - \sin^2 A - 1 + \sin^2 B}{\sin^2 A \sin^2 B} \)

\( = \frac{-\sin^2 A + \sin^2 B}{\sin^2 A \sin^2 B} \)

\( = \frac{\sin^2 B}{\sin^2 A \sin^2 B} - \frac{\sin^2 A}{\sin^2 A \sin^2 B} \)

\( = \frac{1}{\sin^2 A} - \frac{1}{\sin^2 B} \)

\( = \csc^2 A - \csc^2 B \)
In simple words: We convert cot to cos/sin, find common denominator, and use identities to simplify. Finally we get csc²A - csc²B.

📝 Teacher's Note: This involves difference of fractions. Find common denominator first. Use \( \sin^2 + \cos^2 = 1 \) to switch between sin and cos terms.

🎯 Exam Tip: Convert cot to cos/sin first. Show the common denominator step clearly. Final answer is csc²A - csc²B.

Question 12. If \( 4\cos^2A - 3 = 0 \) and \( 0° ≤ A ≤ 90° \), then prove that:
(i) \( \sin 3A = 3 \sin A - 4 \sin^3 A \)
(ii) \( \cos 3A = 4 \cos^3 A - 3 \cos A \)
Answer:
Given: \( 4\cos^2A - 3 = 0 \)

\( \cos A = \frac{\sqrt{3}}{2} \)

We know \( \cos 30° = \frac{\sqrt{3}}{2} \)

So, A = 30°

(i)
LHS = \( \sin 3A = \sin 90° = 1 \)

RHS = \( 3 \sin A - 4 \sin^3 A \)
\( = 3 \sin 30° - 4 \sin^3 30° \)
\( = 3 \times \frac{1}{2} - 4 \times \left(\frac{1}{2}\right)^3 \)
\( = \frac{3}{2} - \frac{1}{2} \)
= 1

LHS = RHS

(ii)
LHS = \( \cos 3A = \cos 90° = 0 \)

RHS = \( 4 \cos^3 A - 3 \cos A \)
\( = 4 \cos^3 30° - 3 \cos 30° \)
\( = 4 \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \left(\frac{\sqrt{3}}{2}\right) \)
\( = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 \)

LHS = RHS
In simple words: We first find A = 30° from the given condition. Then we check both formulas by putting A = 30° and seeing that both sides are equal.

📝 Teacher's Note: Show students how to solve the quadratic equation for cos A first. Then substitute A = 30° in both identities. These are triple angle formulas.

🎯 Exam Tip: First solve for A from the given condition. Write A = 30° clearly. Then substitute in both formulas and show LHS = RHS for each part.

Question 13. Find A, if \( 0° ≤ A ≤ 90° \) and:
(i) \( 2\cos^2 A - 1 = 0 \)
(ii) \( \sin 3A - 1 = 0 \)
(iii) \( 4 \sin^2 A - 3 = 0 \)
(iv) \( \cos^2 A - \cos A = 0 \)
(v) \( 2\cos^2 A + \cos A - 1 = 0 \)
Answer:

(i) \( 2\cos^2 A - 1 = 0 \)

\( \implies \cos^2 A = \frac{1}{2} \)

\( \implies \cos A = \frac{1}{\sqrt{2}} \)

We know \( \cos 45° = \frac{1}{\sqrt{2}} \)

Hence, A = 45°

(ii) \( \sin 3A - 1 = 0 \)
\( \implies \sin 3A = 1 \)
We know \( \sin 90° = 1 \)
\( \therefore 3A = 90° \)
Hence, A = 30°

(iii) \( 4 \sin^2 A - 3 = 0 \)

\( \implies \sin^2 A = \frac{3}{4} \)

\( \implies \sin A = \frac{\sqrt{3}}{2} \)

We know \( \sin 60° = \frac{\sqrt{3}}{2} \)

Hence, A = 60°

(iv) \( \cos^2 A - \cos A = 0 \)
\( \implies \cos A(\cos A - 1) = 0 \)
\( \implies \cos A = 0 \) or \( \cos A = 1 \)
We know \( \cos 90° = 0 \) and \( \cos 0° = 1 \)
Hence, A = 90° or 0°
In simple words: We solve each equation by finding the value of the trigonometric ratio, then finding which angle gives that value.

📝 Teacher's Note: Students must memorize standard angles: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, and corresponding values for cos. Use factoring for quadratic equations.

🎯 Exam Tip: Always solve for the trigonometric ratio first, then find the angle. Know standard angles by heart. For quadratic equations, factor them properly.

Question 14. If 0° < A < 90°; find A, if:
(i) \( \frac{\cos A}{1 - \sin A} + \frac{\cos A}{1 + \sin A} = 4 \)

(ii) \( \frac{\sin A}{\sec A - 1} + \frac{\sin A}{\sec A + 1} = 2 \)

Solution:

(i) \( \frac{\cos A}{1 - \sin A} + \frac{\cos A}{1 + \sin A} = 4 \)

\( \Rightarrow \frac{\cos A + \cos A \sin A + \cos A - \sin A \cos A}{(1 - \sin A)(1 + \sin A)} = 4 \)

\( \Rightarrow \frac{2\cos A}{1 - \sin^2 A} = 4 \)

\( \Rightarrow \frac{2\cos A}{\cos^2 A} = 4 \)

\( \Rightarrow \frac{1}{\cos A} = 2 \)

\( \Rightarrow \cos A = \frac{1}{2} \)

We know \( \cos 60° = \frac{1}{2} \)

Hence, \( A = 60° \)

(ii) \( \frac{\sin A}{\sec A - 1} + \frac{\sin A}{\sec A + 1} = 2 \)

\( \Rightarrow \frac{\sin A \sec A + \sin A + \sec A \sin A - \sin A}{(\sec A - 1)(\sec A + 1)} = 2 \)

\( \Rightarrow \frac{2\sin A \sec A}{\sec^2 A - 1} = 2 \)

\( \Rightarrow \frac{\sin A \sec A}{\tan^2 A} = 1 \)

\( \Rightarrow \frac{\cos A}{\sin A} = 1 \)

\( \Rightarrow \cot A = 1 \)

We know \( \cot 45° = 1 \)

Hence, \( A = 45° \)

In simple words: We used algebra to simplify the fractions. Then we found which angle has those cosine and cotangent values.

📝 Teacher's Note: Show students how to find common denominator first. The key identity is \( 1 - \sin^2 A = \cos^2 A \) and \( \sec^2 A - 1 = \tan^2 A \).

🎯 Exam Tip: Always simplify step by step. Write the final answer clearly as "A = 60°" or "A = 45°". Show all working to get full marks.

Question 15. Prove that: \( (\cosec A - \sin A)(\sec A - \cos A) \sec^2 A = \tan A \)

Solution:

L.H.S.,
\( (\cos \sec A - \sin A)(\sec A - \cos A) \sec^2 A \)

\( = \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) \sec^2 A \)

\( = \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right) \sec^2 A \)

\( = \left(\frac{\cos^2 A}{\sin A}\right)\left(\frac{\sin^2 A}{\cos A}\right) \sec^2 A \)

\( = \frac{\sin A}{\cos A} = \tan A = \text{R.H.S.} \)

In simple words: We changed cosec and sec to fractions. Then we used the identity \( 1 - \sin^2 A = \cos^2 A \) to simplify. Finally we got tan A.

📝 Teacher's Note: Remind students that \( \cosec A = \frac{1}{\sin A} \) and \( \sec A = \frac{1}{\cos A} \). The key step is using \( 1 - \sin^2 A = \cos^2 A \).

🎯 Exam Tip: Write "L.H.S." and "R.H.S." clearly. Show each step of simplification. Always end with "= R.H.S." to show the proof is complete.

Question 16. Prove the identity \( (\sin θ + \cos θ)(\tan θ + \cot θ) = \sec θ + \cosec θ \)

Solution:

L.H.S. = \( (\sin θ + \cos θ)(\tan θ + \cot θ) \)

\( = (\sin θ + \cos θ)\left(\frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ}\right) \)

\( = (\sin θ + \cos θ)\left(\frac{\sin^2 θ + \cos^2 θ}{\cos θ \sin θ}\right) \)

\( = \frac{\sin θ + \cos θ}{\cos θ \sin θ} \)

\( = \frac{\sin θ}{\cos θ \sin θ} + \frac{\cos θ}{\cos θ \sin θ} \)

\( = \frac{1}{\cos θ} + \frac{1}{\sin θ} \)

\( = \sec θ + \cosec θ \)

\( = \text{R.H.S.} \)

In simple words: We wrote tan and cot as fractions. Then we used \( \sin^2 θ + \cos^2 θ = 1 \) to make it simple. Finally we got sec + cosec.

📝 Teacher's Note: The key step is recognizing \( \sin^2 θ + \cos^2 θ = 1 \). Students often forget this basic identity. Practice it separately first.

🎯 Exam Tip: Always convert tan and cot to sin/cos fractions first. This makes the algebra much easier. Write each step clearly.

Question 17. Evaluate without using trigonometric tables, \( \sin^2 28° + \sin^2 62° + \tan^2 38° - \cot^2 52° + \sec^2 30° \)

Solution:

\( \sin^2 28° + \sin^2 62° + \tan^2 38° - \cot^2 52° + \frac{1}{4} \sec^2 30° \)

\( = \sin^2 28° + [\sin (90 - 28)]^2 + \tan^2 38° - [\cot(90 - 38)]^2 + \frac{1}{4} \sec^2 30° \)

\( = \sin^2 28° + \cos^2 28° + \tan^2 38° - \tan^2 38° + \frac{1}{4} \sec^2 30° \)

\( = 1 + 0 + \frac{1}{4} \times \left(\frac{2}{\sqrt{3}}\right)^2 \)

\( = 1 + \frac{1}{3} \)

\( = \frac{3 + 1}{3} \)

\( = \frac{4}{3} \)

In simple words: We used complementary angles. 62° = 90° - 28°, so sin 62° = cos 28°. Also 52° = 90° - 38°, so cot 52° = tan 38°. These pairs cancel out to give us 1.

📝 Teacher's Note: Teach students that sin(90° - A) = cos A and cot(90° - A) = tan A. These complementary angle relationships are very useful for simplifying expressions.

🎯 Exam Tip: Look for complementary angle pairs first. Remember \( \sec 30° = \frac{2}{\sqrt{3}} \). Show all substitutions clearly to get full marks.